#include <cstdio>
#include <algorithm>
using namespace std;
pair<pair<int, int>, int> p[1010];
long long dp[20010];
int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		int w, s, v;
		scanf("%d%d%d", &w, &s, &v);
		p[i] = make_pair(make_pair(s + w, w), v);
	}
	sort(p, p + n);
	long long ans = 0;
	for (int i = 0; i < n; i++) {
		int s = p[i].first.first, w = p[i].first.second, v = p[i].second;
		s -= w;
		for (int j = s; j >= 0; j--) {
			dp[j+w] = max(dp[j+w], dp[j] + v);
		}
		for (int j = 20000; j >= 0; j--) {
			ans = max(ans, dp[j]);
		}
	}
	printf("%lld\n", ans);
	return 0;
}

### Prompt

Please formulate 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>
using namespace std;
pair<pair<int, int>, int> p[1010];
long long dp[20010];
int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		int w, s, v;
		scanf("%d%d%d", &w, &s, &v);
		p[i] = make_pair(make_pair(s + w, w), v);
	}
	sort(p, p + n);
	long long ans = 0;
	for (int i = 0; i < n; i++) {
		int s = p[i].first.first, w = p[i].first.second, v = p[i].second;
		s -= w;
		for (int j = s; j >= 0; j--) {
			dp[j+w] = max(dp[j+w], dp[j] + v);
		}
		for (int j = 20000; j >= 0; j--) {
			ans = max(ans, dp[j]);
		}
	}
	printf("%lld\n", ans);
	return 0;
}

```