#include <iostream>
#include <vector>
// 定义 LL 为 long long 的别名
using LL = long long;
using namespace std;
vector<LL> a;
vector<LL> b;

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

// 中国剩余定理
LL CRT(int n, vector<LL> a, vector<LL> b) {
    LL M = 1;
    // 计算 M 的过程 -> M = a1 * a2 * a3 * …… * an
    for (int i = 0; i < a.size(); i++) {
        M *= a[i];
    }
    LL x = 0;
    for (int i = 0; i < n; ++i) {
        LL Mi = M / a[i];// 计算n个 Mi -> Mi = M/ai
        LL Mi_ni, _;
        // 求 Mi 在模 ai 下的逆元 Mi_ni，即找到 Mi_ni 使得 Mi*Mi_ni ≡ 1(mod ai)
        LL d = exgcd(Mi, a[i], Mi_ni, _);
        //Mi_ni = Mi_ni * 1 / d;
        x += b[i] * Mi * Mi_ni;
    }
    return (x % M + M) % M;  // 确保结果为非负
}

int main() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        int a_, b_;
        cin >> a_ >> b_;
        a.push_back(a_);
        b.push_back(b_);
    }
    cout << CRT(n, a, b) << endl;
    return 0;
}