#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define ll long long
#define N 1010
#define M 10000010
using namespace std;

struct block{ ll w,s,v; };

int n;
ll ans,dp[M];
block bs[N];

bool cmp(block a, block b) { return (a.w+a.s<b.w+b.s); }

int main()
{
    cin>>n; FOR(i,0,n) cin>>bs[i].w>>bs[i].s>>bs[i].v;
    sort(bs,bs+n,cmp);
    FOR(i,0,n)
    {
        for(int j=bs[i].s;j>=0;j--)
        {
            dp[j+bs[i].w]=max(dp[j+bs[i].w],dp[j]+bs[i].v);
        }
    }
    FOR(i,0,M) ans=max(ans,dp[i]);    
    cout<<ans<<'\n';

    return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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>
#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define ll long long
#define N 1010
#define M 10000010
using namespace std;

struct block{ ll w,s,v; };

int n;
ll ans,dp[M];
block bs[N];

bool cmp(block a, block b) { return (a.w+a.s<b.w+b.s); }

int main()
{
    cin>>n; FOR(i,0,n) cin>>bs[i].w>>bs[i].s>>bs[i].v;
    sort(bs,bs+n,cmp);
    FOR(i,0,n)
    {
        for(int j=bs[i].s;j>=0;j--)
        {
            dp[j+bs[i].w]=max(dp[j+bs[i].w],dp[j]+bs[i].v);
        }
    }
    FOR(i,0,M) ans=max(ans,dp[i]);    
    cout<<ans<<'\n';

    return 0;
}
```