#include <bits/stdc++.h>
#define INFLL (0x3f3f3f3f3f3f3f3fll)
using namespace std;
typedef long long ll;

const int MAXN = 1055;

ll dp[20055];

int A[MAXN], B[MAXN], C[MAXN], O[MAXN];

int N;

int main() {
	ios::sync_with_stdio(false);

	cin >> N;
	for(int i = 1; i <= N; i++) cin >> A[i] >> B[i] >> C[i];
	iota(O, O+N+1, 0); sort(O+1, O+N+1, [&](int a, int b) {
		return A[a]+B[a] < A[b]+B[b];
	});

	fill(dp+1, dp+20055, -INFLL);
	for(int oi = 1, i; oi <= N; oi++) {
		i = O[oi];
		for(int j = B[i]; 0 <= j; j--) {
			ll t = dp[j] + C[i];
			if(dp[j+A[i]] < t) dp[j+A[i]] = t;
		}
	}

	cout << *max_element(dp, dp+20055) << endl;
	return 0;
}
### 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 <bits/stdc++.h>
#define INFLL (0x3f3f3f3f3f3f3f3fll)
using namespace std;
typedef long long ll;

const int MAXN = 1055;

ll dp[20055];

int A[MAXN], B[MAXN], C[MAXN], O[MAXN];

int N;

int main() {
	ios::sync_with_stdio(false);

	cin >> N;
	for(int i = 1; i <= N; i++) cin >> A[i] >> B[i] >> C[i];
	iota(O, O+N+1, 0); sort(O+1, O+N+1, [&](int a, int b) {
		return A[a]+B[a] < A[b]+B[b];
	});

	fill(dp+1, dp+20055, -INFLL);
	for(int oi = 1, i; oi <= N; oi++) {
		i = O[oi];
		for(int j = B[i]; 0 <= j; j--) {
			ll t = dp[j] + C[i];
			if(dp[j+A[i]] < t) dp[j+A[i]] = t;
		}
	}

	cout << *max_element(dp, dp+20055) << endl;
	return 0;
}
```