#include <bits/stdc++.h> // Tomasz Nowak
using namespace std;     // XIII LO Szczecin
 
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);

	int n;
	cin >> n;
	vector<int> w(n), s(n), f(n);
	for(int i = 0; i < n; ++i)
		cin >> w[i] >> s[i] >> f[i];

	vector<int> perm(n);
	iota(perm.begin(), perm.end(), 0);
	sort(perm.begin(), perm.end(), [&](int l, int r) {
		return s[l] + w[l] < s[r] + w[r];
	});

	int mxWeight = *max_element(s.begin(), s.end()) + *max_element(w.begin(), w.end());
	vector<long long> dp(mxWeight + 1);

	for(int i = 0; i < n; ++i) {
		int p = perm[i];
		for(int weight = s[p]; weight >= 0; --weight)
			dp[weight + w[p]] = max(dp[weight + w[p]], dp[weight] + f[p]);
	}
	cout << *max_element(dp.begin(), dp.end()) << '\n';
}

### 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> // Tomasz Nowak
using namespace std;     // XIII LO Szczecin
 
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);

	int n;
	cin >> n;
	vector<int> w(n), s(n), f(n);
	for(int i = 0; i < n; ++i)
		cin >> w[i] >> s[i] >> f[i];

	vector<int> perm(n);
	iota(perm.begin(), perm.end(), 0);
	sort(perm.begin(), perm.end(), [&](int l, int r) {
		return s[l] + w[l] < s[r] + w[r];
	});

	int mxWeight = *max_element(s.begin(), s.end()) + *max_element(w.begin(), w.end());
	vector<long long> dp(mxWeight + 1);

	for(int i = 0; i < n; ++i) {
		int p = perm[i];
		for(int weight = s[p]; weight >= 0; --weight)
			dp[weight + w[p]] = max(dp[weight + w[p]], dp[weight] + f[p]);
	}
	cout << *max_element(dp.begin(), dp.end()) << '\n';
}

```