裸题，上模版，，嘿嘿
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
#define ll __int64
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
//求一组解(x,y)使得 ax+by = gcd(a,b), 且|x|+|y|最小(注意求出的 x,y 可能为0或负数)。
//以下代码中d = gcd(a,b)
//能够扩展成求等式 ax+by = c,但c必须是d的倍数才有解,即 (c%gcd(a,b))==0
void extend_gcd (ll a , ll b , ll& d, ll &x , ll &y) {
if(!b){d = a; x = 1; y = 0;}
else {extend_gcd(b, a%b, d, y, x); y-=x*(a/b);}
}
ll inv(ll a, ll n) { //计算%n下 a的逆。假设不存在逆return -1;
ll d, x, y;
extend_gcd(a, n, d, x, y);
return d == 1 ? (x+n)%n : -1;
}
ll work(ll l, ll r, ll *m, ll *a){ //下标[l,r] 方程x%m=a;
ll lcm = 1;
for(ll i = l; i <= r; i++)lcm = lcm/gcd(lcm,m[i])*m[i];
for(ll i = l+1; i <= r; i++) {
ll A = m[l], B = m[i], d, k1, k2, c = a[i]-a[l];
extend_gcd(A,B,d,k1,k2);
if(c%d)return -1;
ll mod = m[i]/d;
ll K = ((k1*c/d)%mod+mod)%mod;
a[l] = m[l]*K + a[l];
m[l] = m[l]*m[i]/d;
}
if(a[l]==0)return lcm;
return a[l];
}
#define N 10005
ll n[N],b[N],len;
int main(){
ll len;
while(cin>>len){
for(ll i = 1; i <= len; i++) cin>>n[i]>>b[i];
cout<<work(1,len,n,b)<<endl;
}
return 0;
}