#include<bits/stdc++.h>
using namespace std;
using LL=long long;
    
struct item{
    int w,s,v;
}a[3005];
    
int n;  LL dp[20005];
    
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d%d%d",&a[i].w,&a[i].s,&a[i].v);
    
    sort(a+1,a+n+1,[&](item &x, item &y){return x.s+x.w<y.s+y.w;});

    for(int i=1;i<=n;i++)
        for(int j=min(20000,a[i].w+a[i].s);j>=a[i].w;j--)
            dp[j]=max(dp[j],dp[j-a[i].w]+a[i].v);
            
    printf("%lld",*max_element(dp+1,dp+20001));
    return 0;
}
### Prompt

Create a solution in CPP for 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;
using LL=long long;
    
struct item{
    int w,s,v;
}a[3005];
    
int n;  LL dp[20005];
    
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d%d%d",&a[i].w,&a[i].s,&a[i].v);
    
    sort(a+1,a+n+1,[&](item &x, item &y){return x.s+x.w<y.s+y.w;});

    for(int i=1;i<=n;i++)
        for(int j=min(20000,a[i].w+a[i].s);j>=a[i].w;j--)
            dp[j]=max(dp[j],dp[j-a[i].w]+a[i].v);
            
    printf("%lld",*max_element(dp+1,dp+20001));
    return 0;
}
```