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)

#include <bits/stdc++.h>
#define FI first
#define SE second
#define EPS 1e-9
#define ALL(a) a.begin(),a.end()
#define SZ(a) int((a).size())
#define MS(s, n) memset(s, n, sizeof(s))
#define FOR(i,a,b) for (int i = (a); i <= (b); i++)
#define FORE(i,a,b) for (int i = (a); i >= (b); i--)
#define FORALL(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define WHATIS(x) cout << #x << " is " << x << endl;
#define ERROR(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
//__builtin_ffs(x) return 1 + index of least significant 1-bit of x
//__builtin_clz(x) return number of leading zeros of x
//__builtin_ctz(x) return number of trailing zeros of x

using namespace std;
using ll = long long;
using ld = double;
typedef pair<int, int> II;
typedef pair<II, int> III;
typedef complex<ld> cd;
typedef vector<cd> vcd;

void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cout << *it << " = " << a << endl;
    err(++it, args...);
}

const ll MODBASE = 1000000007LL;
const int MAXN = 1010;
const int MAXM = 10010;
const int MAXK = 110;
const int MAXQ = 200010;

struct Block {
    int s, w, v;
    Block(int s = 0, int w = 0, int v = 0) : s(s), w(w), v(v) {}
};

bool operator< (Block a, Block b) {
    if (a.s + a.w == b.s + b.w) return a.w < b.w;
    return a.s + a.w < b.s + b.w;
}

int n;
ll dp[MAXN][MAXM];
Block a[MAXN];

int main() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    cin >> n;
    FOR(i,1,n) cin >> a[i].w >> a[i].s >> a[i].v;
    sort(a+1, a+n+1);
    FOR(i,0,n)
        FOR(j,0,10000) dp[i][j] = -1e18;
    // FOR(i,1,n) {
    //     cout << a[i].s << ' ' << a[i].w << ' ' << a[i].v << endl;
    // }
    dp[0][0] = 0;
    FOR(i,1,n) {
        FOR(j,0,10000) {
            dp[i][j] = dp[i-1][j];
            if (j >= a[i].w && j - a[i].w <= a[i].s) dp[i][j] = max(dp[i][j], dp[i-1][j-a[i].w] + a[i].v);
        }
    }
    ll res = 0;
    FOR(i,0,n)
        FOR(j,0,10000) res = max(res, dp[i][j]);
    cout << res;
    return 0;
}