/*
数据不清空，爆零两行泪。
多测不读完，爆零两行泪。
边界不特判，爆零两行泪。
贪心不证明，爆零两行泪。
D P 顺序错，爆零两行泪。
大小少等号，爆零两行泪。
变量不统一，爆零两行泪。
越界不判断，爆零两行泪。
调试不注释，爆零两行泪。
溢出不 l l，爆零两行泪。
*/
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=1000,W_I=10000,S_I=10000;
int n;
struct object{int w,s,v;}obj[N+1];
bool operator<(object x,object y){return y.s-x.w>x.s-y.w;}
int dp[N+1][W_I+S_I+1];
signed main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>obj[i].w>>obj[i].s>>obj[i].v;
	sort(obj+1,obj+n+1);
	for(int i=1;i<=n;i++)for(int j=0;j<=W_I+S_I;j++){
		dp[i][j]=dp[i-1][j];
		if(j>=obj[i].w)dp[i][j]=max(dp[i][j],dp[i-1][min(j-obj[i].w,obj[i].s)]+obj[i].v);
	}
	cout<<dp[n][W_I+S_I];
	return 0;
}
### Prompt

Create a solution in cpp for 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
/*
数据不清空，爆零两行泪。
多测不读完，爆零两行泪。
边界不特判，爆零两行泪。
贪心不证明，爆零两行泪。
D P 顺序错，爆零两行泪。
大小少等号，爆零两行泪。
变量不统一，爆零两行泪。
越界不判断，爆零两行泪。
调试不注释，爆零两行泪。
溢出不 l l，爆零两行泪。
*/
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=1000,W_I=10000,S_I=10000;
int n;
struct object{int w,s,v;}obj[N+1];
bool operator<(object x,object y){return y.s-x.w>x.s-y.w;}
int dp[N+1][W_I+S_I+1];
signed main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>obj[i].w>>obj[i].s>>obj[i].v;
	sort(obj+1,obj+n+1);
	for(int i=1;i<=n;i++)for(int j=0;j<=W_I+S_I;j++){
		dp[i][j]=dp[i-1][j];
		if(j>=obj[i].w)dp[i][j]=max(dp[i][j],dp[i-1][min(j-obj[i].w,obj[i].s)]+obj[i].v);
	}
	cout<<dp[n][W_I+S_I];
	return 0;
}
```