#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

struct Block {
  int w, s, v;
};
bool operator<(Block a, Block b) {
  return a.s+a.w < b.s+b.w;
}

Block a[1000];

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> a[i].w >> a[i].s >> a[i].v;
  sort(a, a+n);
  int maxs = 1e4+1;
  vector<ll> dp(maxs,0);
  ll ans = 0;
  for (int i = 0; i < n; i++) {
    vector<ll> ndp = dp;
    for (int j = 0; j <= a[i].s; j++) {
      ll score = dp[j]+a[i].v;
      ans = max(ans, score);
      if (j+a[i].w < maxs)
	ndp[j+a[i].w] = max(ndp[j+a[i].w], score);
    }
    swap(dp, ndp);
  }
  cout << ans << endl;
}

### Prompt

In Cpp, your task is to solve 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>
using namespace std;
typedef long long ll;

struct Block {
  int w, s, v;
};
bool operator<(Block a, Block b) {
  return a.s+a.w < b.s+b.w;
}

Block a[1000];

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> a[i].w >> a[i].s >> a[i].v;
  sort(a, a+n);
  int maxs = 1e4+1;
  vector<ll> dp(maxs,0);
  ll ans = 0;
  for (int i = 0; i < n; i++) {
    vector<ll> ndp = dp;
    for (int j = 0; j <= a[i].s; j++) {
      ll score = dp[j]+a[i].v;
      ans = max(ans, score);
      if (j+a[i].w < maxs)
	ndp[j+a[i].w] = max(ndp[j+a[i].w], score);
    }
    swap(dp, ndp);
  }
  cout << ans << endl;
}

```