시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
10 초 1024 MB 3 3 3 100.000%

## 문제

First, you are given the positive integer $MOD$.

You have a knapsack, that is empty at first.

You have to perform $Q$ queries.

• In each query, you have first to perform either ADD or REMOVE operation, and then perform FIND.
• For ADD operation, you are given positive integers $w$ and $v$. You put the cookie with weight $w$ and value $v$ to the knapsack.
• For REMOVE operation, you take out the cookie with smallest weight from the knapsack and eat it.
• After each ADD or REMOVE you perform FIND operation: given positive integers $l$ and $r$, you answer the follow question: can you choose cookies from that knapsack so that the $l \leq (X \bmod MOD) \leq r$ ($X$ is sum of the weight of the selected cookies)?. If you can't, output $-1$. Otherwise output the maximum sum of the value of the selected cookies.

## 입력

Input is given in the following format:

$MOD$

$Q$

$t'_1$ $w'_1$ $v'_1$ $l'_1$ $r'_1$

$t'_2$ $w'_2$ $v'_2$ $l'_2$ $r'_2$

$\ldots$

$t'_Q$ $w'_Q$ $v'_Q$ $l'_Q$ $r'_Q$

Queries are encrypted as described later. You can get $t_i, w_i, v_i, l_i, r_i$ by decryption $t'_i, w'_i, v'_i, l'_i, r'_i$.

You may assume that $1 \leq w_i, v_i \leq 10^9$, $0 \leq l_i \leq r_i \leq MOD-1$, $t_i=1$ for ADD+FIND query, $t_i=2$ for REMOVE+FIND query (and in this case $w_i=v_i=0$), that the cookie, which is given by ADD, is heavier than any cookies which added by the previous ADD, and that when executing REMOVE, the knapsack isn't empty.

We prepared the decryption code with C++11 (or later), Java, D, C#. Use class Crypto for decryption. The code of class Crypto and examples of its usage can be uploaded from crypto.zip for C++11 (or later), Java, D, C#.

Here is the example for C++:

#include <cstdint> //uint8_t, uint32_t

class Crypto {
public:
Crypto() {
sm = cnt = 0;
seed();
}

int decode(int z) {
z ^= next();
z ^= (next() << 8);
z ^= (next() << 16);
z ^= (next() << 22);
return z;
}

void query(long long z) {
const long long B = 425481007;
const long long MD = 1000000007;
cnt++;
sm = ((sm * B % MD + z) % MD + MD) % MD;
seed();
}
private:
long long sm;
int cnt;

uint8_t data[256];
int I, J;

void swap_data(int i, int j) {
uint8_t tmp = data[i];
data[i] = data[j];
data[j] = tmp;
}

void seed() {
uint8_t key[8];
for (int i = 0; i < 4; i++) {
key[i] = (sm >> (i * 8));
}
for (int i = 0; i < 4; i++) {
key[i+4] = (cnt >> (i * 8));
}

for (int i = 0; i < 256; i++) {
data[i] = i;
}
I = J = 0;

int j = 0;
for (int i = 0; i < 256; i++) {
j = (j + data[i] + key[i%8]) % 256;
swap_data(i, j);
}
}

uint8_t next() {
I = (I+1) % 256;
J = (J + data[I]) % 256;
swap_data(I, J);
return data[(data[I] + data[J]) % 256];
}
};

The decryption process works in the next way:

• First, you make the instance of class Crypto.
• For each query first call the decode function in order of $t', w', v', l', r'$. The return values are $t, w, v, l, r$. Then perform the query and call the query function with the result of the FIND.

The sample C++ code:

#include <cstdio>
#include <cstdlib>
#include <cstdint> //uint8_t, uint32_t

class Crypto {
...
};

int main() {
int MOD, Q;
scanf("%d %d", &MOD, &Q);
Crypto c;
for (int i = 0; i < Q; i++) {
int t, w, v, l, r;
scanf("%d %d %d %d %d", &t, &w, &v, &l, &r);
t = c.decode(t);
w = c.decode(w);
v = c.decode(v);
l = c.decode(l);
r = c.decode(r);
if (t == 1) {
} else {
(delete candy)
}
long long ans = (answer for query(l, r));
c.query(ans);
printf("%lld\n", ans);
}
}

Note that class Crypto consume time about $200$ to process $Q = 100\,000$.

## 출력

For each query print the result of FIND operation.

## 제한

$0 \leq t'_i, w'_i, v'_i, l'_i, r'_1 \leq 2^{30} - 1$, $2 \leq MOD \leq 500$, $1 \leq Q \leq 100\,000$.

## 예제 입력 1

10
7
281614559 249378726 433981056 466775634 683612866
727071329 787572584 591471796 328464426 757737734
279580343 240336097 538846427 808491898 224313807
222498984 42804452 371605808 667115067 791865961
68683864 1045549765 515479514 1067782238 349547144
907343711 381772625 149003422 879314974 953881571
883899098 700164610 414212891 752949213 972845634


## 예제 출력 1

10
0
-1
21
-1
11
111


## 예제 입력 2

7
20
281614559 249378726 433981094 466775639 683612870
59536386 999828879 241246766 434670565 174365647
172060134 848462699 857413429 182122460 807914643
808426426 600772095 829463884 974102196 354283529
370037909 1024921880 664216868 194331103 140834169
917331875 242953442 205247688 335469789 1055568137
823475244 641321246 617915164 160300810 1073617378
892669150 939175632 904628449 606339993 1059849410
829170894 436718235 288920513 228195002 55212938
772189413 373108543 94133155 610930061 513937768
986619331 175674265 812546186 865335970 605634588
880196843 1071068047 723408215 587598264 380801783
393196081 141080294 584230885 135343295 661927186
5740819 967233824 22597607 888639499 467454437
365679801 515258603 989059385 962028117 761163096
357270919 737051059 569528959 935653628 70506031
869282414 947492121 280522456 96822010 856514221
155948699 826430734 291243254 381421299 617876780
980891674 833928389 1048677341 522527723 223764850
50617939 963598173 281959650 499436870 47455938


## 예제 출력 2

0
134
90
158
-1
22
238
269
179
189
121
53
41
41
-1
58
-1
84
-1
149


## 힌트

The result of decoding Sample 1:

10
7
1 5 10 5 5
2 0 0 0 9
1 7 10 2 4
1 12 11 9 9
2 0 0 1 1
1 22 10 2 3
1 32 100 4 4

The result of decoding Sample 2:

7
20
1 5 44 0 1
1 11 90 0 3
2 0 0 3 4
1 18 68 1 6
1 25 32 2 3
1 31 22 2 3
1 32 26 1 5
1 36 31 3 6
2 0 0 2 5
1 43 10 3 6
2 0 0 5 6
2 0 0 3 4
2 0 0 2 4
2 0 0 1 5
2 0 0 3 5
1 49 48 0 4
2 0 0 1 5
1 50 36 0 6
1 56 48 3 5
1 59 17 3 5