#include <iostream>
using namespace std;
const int N = 25 + 5;   //remember to modify the range of the data!!
typedef long long ll;

ll n, m, t;
ll a[N], b[N], c[N];
ll exgcd(ll a, ll b, ll &x, ll &y)
{
   if(b == 0)
   {
       x = 1, y = 0;
       return a;
   }
   ll g = exgcd(b, a % b, y, x);
   y -= a / b * x;
   return g;
}

ll mul(ll a, ll p, ll mod)
{
   ll res = 0;
   while(p > 0)
   {
       if(p & 1)
           res = (res + a) % mod;
       a = (a + a) % mod;
       p >>= 1;
   }
   return res;
}

int main(void)
{
   ll x, y;
   cin >> n;
   m = 1;
   cin >> a[1] >> b[1];
   for(int i = 2; i <= n; i++)
   {
       cin >> a[i] >> b[i]; //a[i] 是模数， b[i]是模后的余数
       ll g = exgcd(a[i - 1], a[i], x, y);
       if((b[i] - b[i - 1]) % g != 0) //无解不能合成一个方程
       {
           cout << -1 << '\n';
           return 0;
       }
       ll k = (b[i] - b[i - 1]) / g, mod = a[i] / g;
       x = (x % mod + mod) % mod; //保证x 是正数
       x = (mul(k, x, mod) + mod) % mod;
       a[i] = a[i - 1] * a[i] / g; //合成后的模数是最小公倍数
       b[i] = (mul(a[i - 1], x, a[i]) + b[i - 1]) % a[i];// 合成后的 x*
   }


   cout << b[n] << '\n';

   return 0;
}