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

struct block {int s, w, v;};
vector<block> blocks;

int cur[30005];
int pre[30005];

signed main(){
	ios_base::sync_with_stdio(false); cin.tie(0);
	
	int n; cin >> n;
	for(int i = 0;i < n;i++){
		int s, w, v; cin >> w >> s >> v;
		blocks.push_back({s,w,v});
	}
	
	sort(blocks.begin(),blocks.end(), [&](block a, block b){ return a.s + a.w > b.s + b.w; });
	
	for(block B : blocks){
		swap(pre, cur);
		for(int i = 0;i <= 30000;i++) cur[i] = pre[i];
		
		for(int i = B.w;i <= 30000;i++){
			int remS = min(B.s, i - B.w);
			cur[remS] = max(cur[remS], pre[i] + B.v);
		}
		cur[B.s] = max(cur[B.s], B.v);
	}
	
	cout << *max_element(cur,cur+30005);
}

### Prompt

Please formulate a CPP solution 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>
#define int long long
using namespace std;

struct block {int s, w, v;};
vector<block> blocks;

int cur[30005];
int pre[30005];

signed main(){
	ios_base::sync_with_stdio(false); cin.tie(0);
	
	int n; cin >> n;
	for(int i = 0;i < n;i++){
		int s, w, v; cin >> w >> s >> v;
		blocks.push_back({s,w,v});
	}
	
	sort(blocks.begin(),blocks.end(), [&](block a, block b){ return a.s + a.w > b.s + b.w; });
	
	for(block B : blocks){
		swap(pre, cur);
		for(int i = 0;i <= 30000;i++) cur[i] = pre[i];
		
		for(int i = B.w;i <= 30000;i++){
			int remS = min(B.s, i - B.w);
			cur[remS] = max(cur[remS], pre[i] + B.v);
		}
		cur[B.s] = max(cur[B.s], B.v);
	}
	
	cout << *max_element(cur,cur+30005);
}

```