#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const int lim = 20000;
int main() {
	int N;
	cin >> N;
	vector<int> W(N), S(N), V(N);
	for (int i = 0; i < N; ++i) {
		cin >> W[i] >> S[i] >> V[i];
	}
	vector<int> p(N);
	for (int i = 0; i < N; ++i) {
		p[i] = i;
	}
	sort(p.begin(), p.end(), [&](int i, int j) { return W[i] + S[i] < W[j] + S[j]; });
	vector<long long> dp(lim + 1);
	for (int i = 0; i < N; ++i) {
		for (int j = S[p[i]]; j >= 0; --j) {
			dp[j + W[p[i]]] = max(dp[j + W[p[i]]], dp[j] + V[p[i]]);
		}
	}
	cout << *max_element(dp.begin(), dp.end()) << '\n';
	return 0;
}
### Prompt

Your task is to create 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 <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const int lim = 20000;
int main() {
	int N;
	cin >> N;
	vector<int> W(N), S(N), V(N);
	for (int i = 0; i < N; ++i) {
		cin >> W[i] >> S[i] >> V[i];
	}
	vector<int> p(N);
	for (int i = 0; i < N; ++i) {
		p[i] = i;
	}
	sort(p.begin(), p.end(), [&](int i, int j) { return W[i] + S[i] < W[j] + S[j]; });
	vector<long long> dp(lim + 1);
	for (int i = 0; i < N; ++i) {
		for (int j = S[p[i]]; j >= 0; --j) {
			dp[j + W[p[i]]] = max(dp[j + W[p[i]]], dp[j] + V[p[i]]);
		}
	}
	cout << *max_element(dp.begin(), dp.end()) << '\n';
	return 0;
}
```