#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
int N,w[1010],s[1010];
ll v[1010],dp[1010][20010] = {};
vector<pair<int,int>> u;

int main(){
	cin >> N;
	for(int i=1;i<=N;i++){
		cin >> w[i] >> s[i] >> v[i];
		u.push_back({w[i]+s[i],i});
	}
	sort(u.begin(),u.end());
	ll ans = 0;
	for(int i=1;i<=N;i++){
		int a = u[i-1].second;
		for(int j=0;j<=20010;j++){
			if(j>=w[a] && j-w[a]<=s[a]) dp[i][j] = max(dp[i-1][j],dp[i-1][j-w[a]]+v[a]);
			else dp[i][j] = dp[i-1][j];
			ans = max(ans,dp[i][j]);
		}
	}
	cout << ans << endl;
}
### 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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
int N,w[1010],s[1010];
ll v[1010],dp[1010][20010] = {};
vector<pair<int,int>> u;

int main(){
	cin >> N;
	for(int i=1;i<=N;i++){
		cin >> w[i] >> s[i] >> v[i];
		u.push_back({w[i]+s[i],i});
	}
	sort(u.begin(),u.end());
	ll ans = 0;
	for(int i=1;i<=N;i++){
		int a = u[i-1].second;
		for(int j=0;j<=20010;j++){
			if(j>=w[a] && j-w[a]<=s[a]) dp[i][j] = max(dp[i-1][j],dp[i-1][j-w[a]]+v[a]);
			else dp[i][j] = dp[i-1][j];
			ans = max(ans,dp[i][j]);
		}
	}
	cout << ans << endl;
}
```