#include <bits/stdc++.h>
using namespace std;

#define ll long long

const int mxN=1e3, mxW=2e4, M=1e9+7;
int n, w[mxN], s[mxN], v[mxN], p[mxN];
ll dp[mxW+1];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);

	cin >> n;
	for(int i=0; i<n; ++i)
		cin >> w[i] >> s[i] >> v[i];
	iota(p, p+n, 0);
	sort(p, p+n, [&](const int &i, const int &j) {
		return s[i]+w[i]<s[j]+w[j];
	});
	for(int i=0; i<n; ++i) {
		int a=p[i];
		for(int j=s[a]; j>=0; --j)
			dp[j+w[a]]=max(dp[j]+v[a], dp[j+w[a]]);
	}
	ll ans=0;
	for(int i=0; i<=mxW; ++i)
		ans=max(dp[i], ans);
	cout << ans;
}

### Prompt

Your challenge is to write 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;

#define ll long long

const int mxN=1e3, mxW=2e4, M=1e9+7;
int n, w[mxN], s[mxN], v[mxN], p[mxN];
ll dp[mxW+1];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);

	cin >> n;
	for(int i=0; i<n; ++i)
		cin >> w[i] >> s[i] >> v[i];
	iota(p, p+n, 0);
	sort(p, p+n, [&](const int &i, const int &j) {
		return s[i]+w[i]<s[j]+w[j];
	});
	for(int i=0; i<n; ++i) {
		int a=p[i];
		for(int j=s[a]; j>=0; --j)
			dp[j+w[a]]=max(dp[j]+v[a], dp[j+w[a]]);
	}
	ll ans=0;
	for(int i=0; i<=mxW; ++i)
		ans=max(dp[i], ans);
	cout << ans;
}

```