#include <cstdio>
#include <vector>
#include <algorithm>

using namespace std;

int w[1000];
int s[1000];
int v[1000];
long long dp[10001];

int main() {
    int n, i, j;
    long long ans = 0;
    vector <pair<int, int>> z;
    
    scanf("%d", &n);
    
    for (i = 0; i < n; i++) scanf("%d %d %d", &w[i], &s[i], &v[i]);
    
    for (i = 0; i < n; i++) z.push_back(make_pair(w[i] + s[i], i));
    sort(z.begin(), z.end());
    
    for (i = 0; i < n; i++) {
        int x = z[i].second;
        
        for (j = s[x]; j >= 0; j--) {
            ans = max(ans, dp[j] + v[x]);
            if (j + w[x] <= 10000) dp[j + w[x]] = max(dp[j + w[x]], dp[j] + v[x]);
        }
    }
    
    printf("%lld\n", ans);
    
    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 <cstdio>
#include <vector>
#include <algorithm>

using namespace std;

int w[1000];
int s[1000];
int v[1000];
long long dp[10001];

int main() {
    int n, i, j;
    long long ans = 0;
    vector <pair<int, int>> z;
    
    scanf("%d", &n);
    
    for (i = 0; i < n; i++) scanf("%d %d %d", &w[i], &s[i], &v[i]);
    
    for (i = 0; i < n; i++) z.push_back(make_pair(w[i] + s[i], i));
    sort(z.begin(), z.end());
    
    for (i = 0; i < n; i++) {
        int x = z[i].second;
        
        for (j = s[x]; j >= 0; j--) {
            ans = max(ans, dp[j] + v[x]);
            if (j + w[x] <= 10000) dp[j + w[x]] = max(dp[j + w[x]], dp[j] + v[x]);
        }
    }
    
    printf("%lld\n", ans);
    
    return 0;
}

```