#include <bits/stdc++.h>
using namespace std;
int N;
long long dp[100005],ans;
struct Node
{
	int w,s,v;
}pre[1000005];
int temp(Node x,Node y)
{
	return x.s+x.w<y.s+y.w;
}
int main()
{
	scanf("%d",&N);
	for (int i=1;i<=N;i++)
	  scanf("%d%d%d",&pre[i].w,&pre[i].s,&pre[i].v);
    sort(pre+1,pre+N+1,temp);
    for (int i=1;i<=N;i++)
      for (int j=pre[i].w+pre[i].s;j>=pre[i].w;j--)
      {
        dp[j]=max(dp[j],dp[j-pre[i].w]+pre[i].v);
        ans=max(ans,dp[j]);
      }
    printf("%lld",ans);
    return 0;
}
### Prompt

Your task is to create 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;
int N;
long long dp[100005],ans;
struct Node
{
	int w,s,v;
}pre[1000005];
int temp(Node x,Node y)
{
	return x.s+x.w<y.s+y.w;
}
int main()
{
	scanf("%d",&N);
	for (int i=1;i<=N;i++)
	  scanf("%d%d%d",&pre[i].w,&pre[i].s,&pre[i].v);
    sort(pre+1,pre+N+1,temp);
    for (int i=1;i<=N;i++)
      for (int j=pre[i].w+pre[i].s;j>=pre[i].w;j--)
      {
        dp[j]=max(dp[j],dp[j-pre[i].w]+pre[i].v);
        ans=max(ans,dp[j]);
      }
    printf("%lld",ans);
    return 0;
}
```