#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 100009;
//洛谷P4777 【模板】扩展中国剩余定理（EXCRT）

LL n;             // n组数据
LL mod[N], yu[N]; //每组测试数据中，模记为mod[i],余数记为yu[i]

//龟速乘
LL qmul(LL a, LL k, LL b) {
    LL res = 0;
    while (k) {
        if (k & 1) res = (res + a) % b;
        a = (a + a) % b;
        k >>= 1;
    }
    return res;
}

//扩展欧几里得
LL exgcd(LL a, LL b, LL &x, LL &y) {
    if (!b) {
        x = 1, y = 0;
        return a;
    }
    LL d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}

//最小公倍数
LL lcm(LL a, LL b) {
    LL x, y;
    LL d = exgcd(a, b, x, y);
    return (a / d * b);
}

//扩展中国剩余定理
LL excrt() {
    LL M = mod[1], ans = yu[1];
    // M为前i-1个数的最小公倍数,即lcm(mod[1],mod[2],...,mod[n])
    // ans为前i-1个方程的通解
    LL x, y;
    for (int i = 2; i <= n; i++) {
        LL m = mod[i];                      //当前要合并方程的模
        LL res = (yu[i] - ans % m + m) % m; // r2-r1，之所以%mi,是为了防止余数差出现负数
        LL d = exgcd(M, m, x, y);           // 1、求特解x0; 2、求最大公约数d
        if (res % d) return -1;             //无解的情况
        LL t = qmul(x, res / d, m);         //根据ax+by=1的特解翻番得到方程ax+by=c的特解
        ans += t * M;                       //获得前i个方程的通解 r=r + x*lcm(m1,m2,..)
        M = lcm(M, m);                      //更改M的值,把mi也加到M中去，相当于不断的取所有mi的最小公倍数
        ans = (ans % M + M) % M;            //对最小公倍数取模就是最小的合理解
    }
    return ans;
}

int main() {
    ios::sync_with_stdio(false); //加速cin和cout
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> mod[i] >> yu[i];
    cout << excrt() << endl;
    return 0;
}