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

int64_t dp[20001];
void chmax(int64_t& a, int64_t b){
    a = max(a, b);
}

int main(){
    int N, W[1000], S[1000], V[1000];
    cin >> N;
    for(int i=0; i<N; i++) cin >> W[i] >> S[i] >> V[i];
    vector<int> order;
    for(int i=0; i<N; i++) order.push_back(i);
    sort(order.begin(), order.end(), [&](int i, int j){ return S[i] + W[i] < S[j] + W[j];});

    for(int i=0; i<N; i++){
        int k = order[i];
        for(int j=20000; j>=0; j--){
            if(j<=S[k]) chmax(dp[j+W[k]], dp[j] + V[k]);
        }
    }
    int64_t ans = 0;
    for(int j=0; j<=20000; j++) chmax(ans, dp[j]);
    cout << ans << endl;
    return 0;
}
### Prompt

Please create a solution in CPP 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 <bits/stdc++.h>
using namespace std;

int64_t dp[20001];
void chmax(int64_t& a, int64_t b){
    a = max(a, b);
}

int main(){
    int N, W[1000], S[1000], V[1000];
    cin >> N;
    for(int i=0; i<N; i++) cin >> W[i] >> S[i] >> V[i];
    vector<int> order;
    for(int i=0; i<N; i++) order.push_back(i);
    sort(order.begin(), order.end(), [&](int i, int j){ return S[i] + W[i] < S[j] + W[j];});

    for(int i=0; i<N; i++){
        int k = order[i];
        for(int j=20000; j>=0; j--){
            if(j<=S[k]) chmax(dp[j+W[k]], dp[j] + V[k]);
        }
    }
    int64_t ans = 0;
    for(int j=0; j<=20000; j++) chmax(ans, dp[j]);
    cout << ans << endl;
    return 0;
}
```