#include <iostream>
#include <algorithm>
using namespace std;
struct block {
	int wt, st;
	long long val;
	bool operator < (const block &b) const {
		return wt+st < b.wt+b.st;
	}
};
int n;
block a[1010];
long long dp[20010];
long long o = 0;
int main () {
	cin >> n;
	for (int i=0; i<n; i++) {
		cin >> a[i].wt >> a[i].st >> a[i].val;
	}
	sort(a, a+n);
	for (int i=0; i<n; i++) {
		for (int j=min(a[i].st, 20005-a[i].wt); j>=0; j--) {
			dp[j+a[i].wt] = max(dp[j+a[i].wt], dp[j]+a[i].val);
			o = max(o, dp[j+a[i].wt]);
		}
	}
	cout << o << endl;
}
### 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 <iostream>
#include <algorithm>
using namespace std;
struct block {
	int wt, st;
	long long val;
	bool operator < (const block &b) const {
		return wt+st < b.wt+b.st;
	}
};
int n;
block a[1010];
long long dp[20010];
long long o = 0;
int main () {
	cin >> n;
	for (int i=0; i<n; i++) {
		cin >> a[i].wt >> a[i].st >> a[i].val;
	}
	sort(a, a+n);
	for (int i=0; i<n; i++) {
		for (int j=min(a[i].st, 20005-a[i].wt); j>=0; j--) {
			dp[j+a[i].wt] = max(dp[j+a[i].wt], dp[j]+a[i].val);
			o = max(o, dp[j+a[i].wt]);
		}
	}
	cout << o << endl;
}
```