#include <iostream>
#include <algorithm>
#include <utility>

using namespace std;
long long w[1010],s[1010],v[1010],dp[1010][20010] = {},ans = 0;
pair<pair<int,int>,int> p[1010];
int main(){
	int i,j,n;
	cin >> n;
	for(i=0;i<n;i++){
		cin >> w[i] >> s[i] >> v[i];
		p[i] = {{w[i]+s[i],-v[i]},i};
	}
	sort(p,p+n);
	for(i=1;i<=n;i++){
		int x = p[i-1].second;
		for(j=0;j<=20000;j++){
			dp[i][j] = max(dp[i][j],dp[i-1][j]);
			if(j>s[x] || w[x]+j>20000) continue;
			dp[i][w[x]+j] = max(dp[i][w[x]+j],dp[i-1][j] + v[x]);
		}
	}
	for(i=0;i<=20000;i++){
		ans = max(ans,dp[n][i]);
	}
	cout << ans << endl;
}
### Prompt

Generate 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 <iostream>
#include <algorithm>
#include <utility>

using namespace std;
long long w[1010],s[1010],v[1010],dp[1010][20010] = {},ans = 0;
pair<pair<int,int>,int> p[1010];
int main(){
	int i,j,n;
	cin >> n;
	for(i=0;i<n;i++){
		cin >> w[i] >> s[i] >> v[i];
		p[i] = {{w[i]+s[i],-v[i]},i};
	}
	sort(p,p+n);
	for(i=1;i<=n;i++){
		int x = p[i-1].second;
		for(j=0;j<=20000;j++){
			dp[i][j] = max(dp[i][j],dp[i-1][j]);
			if(j>s[x] || w[x]+j>20000) continue;
			dp[i][w[x]+j] = max(dp[i][w[x]+j],dp[i-1][j] + v[x]);
		}
	}
	for(i=0;i<=20000;i++){
		ans = max(ans,dp[n][i]);
	}
	cout << ans << endl;
}
```