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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pii;
typedef pair<double,double> pdd;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define SZ(i) ll(i.size())
#define FOR(i, j, k, in) for (ll i=j ; i<k ; i+=in)
#define RFOR(i, j, k, in) for (ll i=j ; i>=k ; i-=in)
#define REP(i, j) FOR(i, 0, j, 1)
#define REP1(i,j) FOR(i, 1, j+1, 1)
#define RREP(i, j) RFOR(i, j, 0, 1)
#define ALL(_a) _a.begin(),_a.end()
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define X first
#define Y second
#ifdef tmd
#define debug(...) do{\
    fprintf(stderr,"%s - %d (%s) = ",__PRETTY_FUNCTION__,__LINE__,#__VA_ARGS__);\
    _do(__VA_ARGS__);\
}while(0)
template<typename T>void _do(T &&_x){cerr<<_x<<endl;}
template<typename T,typename ...S> void _do(T &&_x,S &&..._t){cerr<<_x<<" ,";_do(_t...);}
template<typename _a,typename _b> ostream& operator << (ostream &_s,const pair<_a,_b> &_p){return _s<<"("<<_p.X<<","<<_p.Y<<")";}
template<typename It> ostream& _OUTC(ostream &_s,It _ita,It _itb)
{
    _s<<"{";
    for(It _it=_ita;_it!=_itb;_it++)
    {
        _s<<(_it==_ita?"":",")<<*_it;
    }
    _s<<"}";
    return _s;
}
template<typename _a> ostream &operator << (ostream &_s,vector<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a> ostream &operator << (ostream &_s,set<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a,typename _b> ostream &operator << (ostream &_s,map<_a,_b> &_c){return _OUTC(_s,ALL(_c));}
template<typename _t> void pary(_t _a,_t _b){_OUTC(cerr,_a,_b);cerr<<endl;}
#define IOS()
#else
#define debug(...)
#define pary(...)
#define endl '\n'
#define IOS() ios_base::sync_with_stdio(0);cin.tie(0)
#endif

template<class T> inline bool cmax(T &a, const T &b) { return b > a ? a = b, true : false; }
template<class T> inline bool cmin(T &a, const T &b) { return b < a ? a = b, true : false; }
template<class T> using MaxHeap = priority_queue<T>;
template<class T> using MinHeap = priority_queue<T, vector<T>, greater<T>>;

const ll MOD=1000000007;
const ll INF=0x3f3f3f3f3f3f3f3f;
const ll MAXN = 1003;
const ll MAXW = 20003;
const ll MAXLG=__lg(MAXN)+2;
ll n;
struct Block {
    ll w,s,v,id;
}blocks[MAXN];
ll dp[MAXN][MAXW];
/********** Good Luck :) **********/
int main()
{
    IOS();
    cin >> n;
    REP (i,n) {
        cin >> blocks[i].w >> blocks[i].s >> blocks[i].v;
        blocks[i].id = i + 1;
    }

    sort(blocks,blocks+n,[&](const Block &b1,const Block &b2){
        return b1.w + b1.s < b2.w + b2.s;
    });

    MEM(dp,-1);
    
    REP (i,n) {
        if (i == 0) {
            dp[i][0] = 0;
            dp[i][blocks[i].w] = blocks[i].v;
        } else {
            for (ll j=0;j<MAXW;j++) {
                if (j-blocks[i].w >= 0 && j-blocks[i].w <= blocks[i].s && dp[i-1][j-blocks[i].w] != -1) {
                    dp[i][j] = dp[i-1][j-blocks[i].w] + blocks[i].v;
                }
                dp[i][j] = max(dp[i][j],dp[i-1][j]);
            }
        }
    }

    ll ans = 0;
    REP (i,MAXW) {
        ans = max(ans,dp[n-1][i]);
    }

    cout << ans << endl;
    return 0;
}
//DP