#include <bits/stdc++.h>

using namespace std;

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  
  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> ord(n);
  iota(ord.begin(), ord.end(), 0);
  sort(ord.begin(), ord.end(), [&](int i, int j) { return s[i] - w[j] < s[j] - w[i]; });

  const int MAXS = 20010;
  vector<long long> dp(MAXS);
  for (int i = 0; i < n; i++) {
    for (int j = s[ord[i]]; j >= 0; j--) {
      dp[j + w[ord[i]]] = max(dp[j + w[ord[i]]], dp[j] + v[ord[i]]);
    }
  }
  
  cout << *max_element(dp.begin(), dp.end()) << '\n';
  return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>

using namespace std;

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  
  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> ord(n);
  iota(ord.begin(), ord.end(), 0);
  sort(ord.begin(), ord.end(), [&](int i, int j) { return s[i] - w[j] < s[j] - w[i]; });

  const int MAXS = 20010;
  vector<long long> dp(MAXS);
  for (int i = 0; i < n; i++) {
    for (int j = s[ord[i]]; j >= 0; j--) {
      dp[j + w[ord[i]]] = max(dp[j + w[ord[i]]], dp[j] + v[ord[i]]);
    }
  }
  
  cout << *max_element(dp.begin(), dp.end()) << '\n';
  return 0;
}
```