#include <cstdio>
#include <iostream>
using namespace std;
static const int MAX_N = 101;
static const int MAX_W = 10001;
int n, W;
int v[MAX_N], w[MAX_N], m[MAX_N];
int dp[MAX_W];
int deq[MAX_W], deqv[MAX_W];

void solve() {
    for (int i = 0; i < n; i++) {
        for (int a = 0; a < w[i]; a++) {
            int s = 0, t = 0;
            for (int j = 0; j * w[i] + a <= W; j++) {
                int val = dp[j * w[i] + a] - j * v[i];
                while (s < t && deqv[t-1] <= val) t--;
                deq[t] = j;
                deqv[t++] = val;
                dp[j * w[i] + a] = deqv[s] + j * v[i];
                if (deq[s] == j - m[i]) s++;
            }
        }
    }
    printf("%d\n", dp[W]);
}

int main() {
    scanf("%d %d", &n, &W);
    for (int i = 0; i < n; i++) {
        scanf("%d %d %d", &v[i], &w[i], &m[i]);
    }

    //fill(deq, deq+W+1, 0);
    //fill(deqv, deqv+W+1, 0);
    solve();
    return 0;
}


### Prompt

Develop a solution in CPP to the problem described below:

You have N items that you want to put them into a knapsack. Item i has value vi, weight wi and limitation mi.

You want to find a subset of items to put such that:

* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
* You can select at most mi items for ith item.



Find the maximum total value of items in the knapsack.

Constraints

* 1 ≤ N ≤ 100
* 1 ≤ vi ≤ 1,000
* 1 ≤ wi ≤ 1,000
* 1 ≤ mi ≤ 10,000
* 1 ≤ W ≤ 10,000

Input


N W
v1 w1 m1
v2 w2 m2
:
vN wN mN


The first line consists of the integers N and W. In the following N lines, the value, weight and limitation of the i-th item are given.

Output

Print the maximum total values of the items in a line.

Examples

Input

4 8
4 3 2
2 1 1
1 2 4
3 2 2


Output

12


Input

2 100
1 1 100
2 1 50


Output

150

### Response

```cpp
#include <cstdio>
#include <iostream>
using namespace std;
static const int MAX_N = 101;
static const int MAX_W = 10001;
int n, W;
int v[MAX_N], w[MAX_N], m[MAX_N];
int dp[MAX_W];
int deq[MAX_W], deqv[MAX_W];

void solve() {
    for (int i = 0; i < n; i++) {
        for (int a = 0; a < w[i]; a++) {
            int s = 0, t = 0;
            for (int j = 0; j * w[i] + a <= W; j++) {
                int val = dp[j * w[i] + a] - j * v[i];
                while (s < t && deqv[t-1] <= val) t--;
                deq[t] = j;
                deqv[t++] = val;
                dp[j * w[i] + a] = deqv[s] + j * v[i];
                if (deq[s] == j - m[i]) s++;
            }
        }
    }
    printf("%d\n", dp[W]);
}

int main() {
    scanf("%d %d", &n, &W);
    for (int i = 0; i < n; i++) {
        scanf("%d %d %d", &v[i], &w[i], &m[i]);
    }

    //fill(deq, deq+W+1, 0);
    //fill(deqv, deqv+W+1, 0);
    solve();
    return 0;
}


```