#include <bits/stdc++.h>
#define mk make_pair
using namespace std;
typedef long long ll;
const int N = 3e3 + 5;
int n, m, w[N], s[N], v[N], a[N];
ll dp[N][20001], ans;
int main() {
	cin >> n;
	for (int i = 1; i <= n; ++i) {
		scanf("%d %d %d", w + i, s + i, v + i);
		a[i] = i;
	}
	sort(a + 1, a + 1 + n, [&](int i, int j) {
		return s[i] + w[i] < s[j] + w[j];
	});
	for (int i = 1; i <= n; ++i) {
		ans = max(ans, (ll)v[i]);
		dp[i][w[a[i]]] = max(dp[i - 1][w[a[i]]], (ll)v[a[i]]);
		for (int j = 0; j <= 20000; ++j) {
			dp[i][j] = dp[i - 1][j];
			if (j - w[a[i]] >= 0 && j - w[a[i]] <= s[a[i]])
				dp[i][j] = max(dp[i][j], dp[i - 1][j - w[a[i]]] + v[a[i]]);
			ans = max(ans, dp[i][j]);	
		}
	}
	cout << ans;
}


### Prompt

Create a solution in CPP for 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 mk make_pair
using namespace std;
typedef long long ll;
const int N = 3e3 + 5;
int n, m, w[N], s[N], v[N], a[N];
ll dp[N][20001], ans;
int main() {
	cin >> n;
	for (int i = 1; i <= n; ++i) {
		scanf("%d %d %d", w + i, s + i, v + i);
		a[i] = i;
	}
	sort(a + 1, a + 1 + n, [&](int i, int j) {
		return s[i] + w[i] < s[j] + w[j];
	});
	for (int i = 1; i <= n; ++i) {
		ans = max(ans, (ll)v[i]);
		dp[i][w[a[i]]] = max(dp[i - 1][w[a[i]]], (ll)v[a[i]]);
		for (int j = 0; j <= 20000; ++j) {
			dp[i][j] = dp[i - 1][j];
			if (j - w[a[i]] >= 0 && j - w[a[i]] <= s[a[i]])
				dp[i][j] = max(dp[i][j], dp[i - 1][j - w[a[i]]] + v[a[i]]);
			ans = max(ans, dp[i][j]);	
		}
	}
	cout << ans;
}


```