#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;

struct point {
	int a, b, c;
	bool operator<(const point &p)const {
		return a + b < p.a + p.b;
	}
}w[1010];
long long D[20100];
int n;
int main() {
	int i, j;
	scanf("%d", &n);
	for (i = 1; i <= n; i++) {
		scanf("%d%d%d", &w[i].a, &w[i].b, &w[i].c);
	}
	sort(w + 1, w + n + 1);

	for (i = 1; i <= n; i++) {
		for (j = w[i].b; j >= 0; j--) {
			D[j + w[i].a] = max(D[j + w[i].a], D[j] + w[i].c);
		}
	}
	long long res = 0;
	for (i = 0; i <= 20000; i++)res = max(res, D[i]);
	printf("%lld\n", res);
}

### 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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;

struct point {
	int a, b, c;
	bool operator<(const point &p)const {
		return a + b < p.a + p.b;
	}
}w[1010];
long long D[20100];
int n;
int main() {
	int i, j;
	scanf("%d", &n);
	for (i = 1; i <= n; i++) {
		scanf("%d%d%d", &w[i].a, &w[i].b, &w[i].c);
	}
	sort(w + 1, w + n + 1);

	for (i = 1; i <= n; i++) {
		for (j = w[i].b; j >= 0; j--) {
			D[j + w[i].a] = max(D[j + w[i].a], D[j] + w[i].c);
		}
	}
	long long res = 0;
	for (i = 0; i <= 20000; i++)res = max(res, D[i]);
	printf("%lld\n", res);
}

```