Problem:
Score : 100 points

Problem Statement

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.

Sample Input 1

3
2 2 20
2 1 30
3 1 40

Sample Output 1

50

If Blocks 2, 1 are stacked in this order from top to bottom, this tower will satisfy the condition, with the total value of 30 + 20 = 50.

Sample Input 2

4
1 2 10
3 1 10
2 4 10
1 6 10

Sample Output 2

40

Blocks 1, 2, 3, 4 should be stacked in this order from top to bottom.

Sample Input 3

5
1 10000 1000000000
1 10000 1000000000
1 10000 1000000000
1 10000 1000000000
1 10000 1000000000

Sample Output 3

5000000000

The answer may not fit into a 32-bit integer type.

Sample Input 4

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

Sample Output 4

22

We should, for example, stack Blocks 5, 6, 8, 4 in this order from top to bottom.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/* --- author: dxm --- */
#include<bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;

#define INF 1e9
#define Inf 1e18
#define rep(i,n) for(int (i)=0;(i)<n;i++)
#define REP(i,n) for(int (i)=1;(i)<=n;i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef double db;
typedef long double ldb;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;

void qread(int &x){
	int neg=1;x=0;
	char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')neg=-1;c=getchar();}
	while(c>='0'&&c<='9')x=10*x+c-'0',c=getchar();
	x*=neg;
}

const int maxn=1005;
const int maxm=10005;
int n,w[maxn],s[maxn],v[maxn],p[maxn];
ll dp[maxn][maxm],mx[maxm];

bool cmp(int a,int b){
	return w[a]+s[a]<w[b]+s[b];
}

int main(){
#ifdef FILE
	freopen("FILE.in","r",stdin);
	freopen("FILE.out","w",stdout);
#endif
	scanf("%d",&n);
	REP(i,n)scanf("%d%d%d",&w[i],&s[i],&v[i]),p[i]=i;
	sort(p+1,p+1+n,cmp);
	ll ans=0;
	REP(i,n){
		int x=p[i];
		for(int j=w[x];j<=10000;j++){
			if(j-w[x]<=s[x])dp[i][j]=mx[j-w[x]]+v[x];
			ans=max(ans,dp[i][j]);
		}
		for(int j=w[x];j<=10000;j++)
			mx[j]=max(mx[j],dp[i][j]);
	}
	printf("%lld\n",ans);
#ifdef TIME
	printf("Running Time = %d ms\n",int(clock()*1000.0/CLOCKS_PER_SEC));
#endif
	return 0;
}

/*
Input:
-----------------
Output:
*/