#include<bits/stdc++.h>
using namespace std;
const long long maxn=20000;
long long n,w,s,v,dp[1111][22222],ans;
pair<long long,pair<long long,long long> > a[1111];
int main()
{
	scanf("%lld",&n);
	for (long long i=1;i<=n;i++)
	{
		scanf("%lld%lld%lld",&w,&s,&v);
		s+=w;
		a[i]=make_pair(s,make_pair(w,v));
	}
	sort(a+1,a+n+1);
	for (long long i=0;i<=n;i++)
	{
		for (long long j=0;j<=maxn;j++)
		dp[i][j]=-1e18;
	}
	dp[0][0]=0;
	for (long long i=1;i<=n;i++)
	{
		for (long long j=0;j<=maxn;j++)
		{
			dp[i][j]=max(dp[i][j],dp[i-1][j]);
			if (j>=a[i].second.first && j<=a[i].first) dp[i][j]=max(dp[i][j],dp[i-1][j-a[i].second.first]+a[i].second.second);
		}
	}
	for (long long i=0;i<=maxn;i++)
	{
		ans=max(ans,dp[n][i]);
	}
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Generate 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>
using namespace std;
const long long maxn=20000;
long long n,w,s,v,dp[1111][22222],ans;
pair<long long,pair<long long,long long> > a[1111];
int main()
{
	scanf("%lld",&n);
	for (long long i=1;i<=n;i++)
	{
		scanf("%lld%lld%lld",&w,&s,&v);
		s+=w;
		a[i]=make_pair(s,make_pair(w,v));
	}
	sort(a+1,a+n+1);
	for (long long i=0;i<=n;i++)
	{
		for (long long j=0;j<=maxn;j++)
		dp[i][j]=-1e18;
	}
	dp[0][0]=0;
	for (long long i=1;i<=n;i++)
	{
		for (long long j=0;j<=maxn;j++)
		{
			dp[i][j]=max(dp[i][j],dp[i-1][j]);
			if (j>=a[i].second.first && j<=a[i].first) dp[i][j]=max(dp[i][j],dp[i-1][j-a[i].second.first]+a[i].second.second);
		}
	}
	for (long long i=0;i<=maxn;i++)
	{
		ans=max(ans,dp[n][i]);
	}
	printf("%lld\n",ans);
	return 0;
}
```