Problem:
Score : 100 points

Problem Statement

Consider an infinite sequence a_1, a_2, … Initially, the values of all the terms are 0, and from this state we will sequentially perform Q operations. The i-th operation (1 ≤ i ≤ Q) is as follows:

For every positive integer j, add x_i to the value of a_{j × m_i}.

Find the value of the largest term after these Q operations.

Constraints

1 ≤ Q ≤ 299

2 ≤ m_i ≤ 300

-10^6 ≤ x_i ≤ 10^6

All m_i are distinct.

All input values are integers.

Input

Input is given from Standard Input in the following format:

Q
m_1 x_1
:
m_Q x_Q

Output

Print the value of the largest term after the Q operations.

Sample Input 1

3
2 10
3 -20
6 15

Sample Output 1

10

The values of each terms in the sequence a_1, a_2, … change as follows:

Before the operations: 0, 0, 0, 0, 0, 0, …

After the 1-st operation: 0, 10, 0, 10, 0, 10, …

After the 2-nd operation: 0, 10, -20, 10, 0, -10, …

After the 3-rd operation: 0, 10, -20, 10, 0, 5, …

The value of the largest term after all the operations is 10.

Sample Input 2

3
10 -3
50 4
100 -5

Sample Output 2

1

Sample Input 3

5
56 114834
72 -149861
100 190757
192 -132693
240 133108

Sample Output 3

438699

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

#include <bits/stdc++.h>
using namespace std;

struct Dinic{
    struct edge {
        int64_t to, cap, rev;
        edge(int64_t to, int64_t cap, int64_t rev):to(to), cap(cap), rev(rev){}
    };

    int N, S, T;
    int64_t flow = 0;
    const int64_t INF = 1e18;
    vector<int> level, iter;
    vector<vector<edge>> G;

    Dinic(int N, int S, int T):N(N), S(S), T(T){
        flow = 0;
        level.resize(N);
        iter.resize(N);
        G.resize(N);
    }

    void add_edge(int from, int to, int64_t cap){
        G[from].emplace_back(to, cap, G[to].size());
        G[to].emplace_back(from, 0, G[from].size()-1);
    }

    void bfs(int s){
        fill(level.begin(), level.end(), -1);
        queue<int> que;
        level[s] = 0;
        que.push(s);
        while(que.size()){
            int v = que.front(); que.pop();
            for(auto& e : G[v]){
                if(e.cap > 0 && level[e.to] < 0){
                    level[e.to] = level[v] + 1;
                    que.push(e.to);
                }
            }
        }
    }

    int64_t dfs(int v, int t, int64_t f){
        if(v == t) return f;
        for(int& i=iter[v]; i<G[v].size(); i++){
            auto& e = G[v][i];
            if(e.cap > 0 && level[v] < level[e.to]){
                int64_t d = dfs(e.to, t, min(f, e.cap));
                if(d > 0){
                    e.cap -= d;
                    G[e.to][e.rev].cap += d;
                    return d;
                }
            }
        }
        return 0;
    }

    int64_t max_flow(){
        while(true){
            bfs(S);
            if(level[T] < 0) return flow;
            fill(iter.begin(), iter.end(), 0);
            while(true){
                int64_t f = dfs(S, T, INF);
                if(f == 0) break;
                flow += f;
            }
        }
    }
};

int main(){
    int N, M[300], X[300];
    cin >> N;
    for(int i=0; i<N; i++) cin >> M[i] >> X[i];

    int64_t fix = 0;
    int S = 300, T = 301;
    Dinic solver(T+1, S, T);
    for(int i=0; i<N; i++){
        if(X[i] > 0){
            fix += X[i];
            solver.add_edge(S, i, 0);
            solver.add_edge(i, T, X[i]);
        }else{
            solver.add_edge(S, i, -X[i]);
            solver.add_edge(i, T, 0);
        }
    }

    const int64_t INF = 1e15;
    for(int i=0; i<N; i++) for(int j=0; j<N; j++){
        if(i != j && M[i] % M[j] == 0){
            solver.add_edge(j, i, INF);
        }
    }

    int64_t ans = fix - solver.max_flow();
    cout << ans << endl;
}