#include <bits/stdc++.h>

using namespace std;

struct box {
    int w, s, v;
};
int n;
box arr[1010];
long long D[20010];

int main() {
    int i, j;

    scanf("%d",&n);
    for (i=0;i<n;i++) scanf("%d%d%d",&arr[i].w,&arr[i].s,&arr[i].v);
    sort(arr,arr+n,[](box &a, box &b) {return a.w+a.s<b.w+b.s;});
    for (i=0;i<n;i++) {
        for (j=arr[i].w+arr[i].s;j>=arr[i].w;j--) D[j] = max(D[j],D[j-arr[i].w]+arr[i].v);
        for (j=1;j<=20001;j++) D[j] = max(D[j],D[j-1]);
    }
    printf("%lld\n",D[20001]);

    return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

There are N blocks, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), Block i has a weight of w_i, a solidness of s_i and a value of v_i.

Taro has decided to build a tower by choosing some of the N blocks and stacking them vertically in some order. Here, the tower must satisfy the following condition:

* For each Block i contained in the tower, the sum of the weights of the blocks stacked above it is not greater than s_i.



Find the maximum possible sum of the values of the blocks contained in the tower.

Constraints

* All values in input are integers.
* 1 \leq N \leq 10^3
* 1 \leq w_i, s_i \leq 10^4
* 1 \leq v_i \leq 10^9

Input

Input is given from Standard Input in the following format:


N
w_1 s_1 v_1
w_2 s_2 v_2
:
w_N s_N v_N


Output

Print the maximum possible sum of the values of the blocks contained in the tower.

Examples

Input

3
2 2 20
2 1 30
3 1 40


Output

50


Input

4
1 2 10
3 1 10
2 4 10
1 6 10


Output

40


Input

5
1 10000 1000000000
1 10000 1000000000
1 10000 1000000000
1 10000 1000000000
1 10000 1000000000


Output

5000000000


Input

8
9 5 7
6 2 7
5 7 3
7 8 8
1 9 6
3 3 3
4 1 7
4 5 5


Output

22

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;

struct box {
    int w, s, v;
};
int n;
box arr[1010];
long long D[20010];

int main() {
    int i, j;

    scanf("%d",&n);
    for (i=0;i<n;i++) scanf("%d%d%d",&arr[i].w,&arr[i].s,&arr[i].v);
    sort(arr,arr+n,[](box &a, box &b) {return a.w+a.s<b.w+b.s;});
    for (i=0;i<n;i++) {
        for (j=arr[i].w+arr[i].s;j>=arr[i].w;j--) D[j] = max(D[j],D[j-arr[i].w]+arr[i].v);
        for (j=1;j<=20001;j++) D[j] = max(D[j],D[j-1]);
    }
    printf("%lld\n",D[20001]);

    return 0;
}

```