#include <bits/stdc++.h>

#define int  long long

using namespace std;

const int N = 11;

int n;
int M = 1;
int ans = 0;
int phi[N];

int mul(int a, int b)
{
    int res = 0;
    for ( ; b ; b >>= 1)
    {
        if (b & 1) res = (res + a) % M;
        a = (a + a) % M;
    }
    return res % M;
}

int pw(int a, int b)
{
    int res = 1;
    for ( ; b ; b >>= 1)
    {
        if (b & 1) res = mul(res, a) % M;
        a = mul(res, a) % M;
    }

    return res % M;
}

signed main()
{
    scanf("%lld", &n);

    int a[N], m[N];
    for (int i = 1 ; i <= n ; i ++ )
    {
        scanf("%lld%lld", &m[i], &a[i]);
        M *= m[i];
    }

    for (int i = 1 ; i <= n ; i ++ ) // 处理出mi的欧拉函数
    {
        int tmp = phi[i] = m[i];
        for (int j = 2 ; j <= tmp / j ; j ++ )
        {
            if (tmp % j == 0)
            {
                while (tmp % j == 0) tmp /= j;
                phi[i] = phi[i] / j * (j - 1);
            }
        }
        if (tmp != 1) phi[i] = phi[i] / tmp * (tmp - 1);
    }

    for (int i = 1 ; i <= n ; i ++ )
    {
        ans += mul(mul(a[i],  (M / m[i])) % M, pw(M / m[i], phi[i] - 2)); //中国剩余定理公式
        ans %= M;
    }

    printf("%lld", ans);

    return 0;
}