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>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define svec(v) cout<<#v<<":";rep(kbrni,v.size())cout<<" "<<v[kbrni];cout<<endl
#define sset(s) cout<<#s<<":";each(kbrni,s)cout<<" "<<kbrni;cout<<endl
#define smap(m) cout<<#m<<":";each(kbrni,m)cout<<" {"<<kbrni.first<<":"<<kbrni.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 301;

const int p[] = {2,3,5,7,11,13,17};
const int q[] = {9,6,4,3,3,3,3};

bool is_prime[MAX_N];

void sieve(int n){
	for(int i=0;i<=n;i++){
		is_prime[i] = true;
	}
	is_prime[0] = is_prime[1] = false;
	for(int i=2;i<=n;i++){
		if(is_prime[i]){
			for(int j=2*i;j<=n;j+=i){
				is_prime[j] = false;
			}
		}
	}
}

int n;
vi m,x;
int ans;

void dfs(ll val,int id)
{
    if(id == 7){
        int res = 0;
        rep(i,n){
            if(m[i] >= 19 && is_prime[m[i]]){
                res += max(0,x[i]);
            }else if(val % m[i] == 0){
                res += x[i];
            }
        }
        cmx(ans,res);
        return;
    }
    rep(i,q[id]){
        dfs(val*(ll)pow(p[id],i),id+1);
    }
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> n;
    m.resize(n),x.resize(n);
    sieve(300);
    rep(i,n){
        cin >> m[i] >> x[i];
    }
    ans = 0;
    // dfs(1LL,0);
    cout << ans << "\n";
    return 0;
}