#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int (i)=(0);(i)<(int)(n);++(i))
using ll = long long;
using P = pair<ll, ll>;
template<class T> inline bool chmax(T& a, T b) {if (a < b){a = b;return 1;}return 0;}
template<class T> inline bool chmin(T& a, T b) {if (a > b){a = b;return 1;}return 0;}
const long long LLINF = 1LL << 60;
using namespace std;
// 負の数にも対応したmod
//
inline long long mod(long long a, long long m) {
    return (a % m + m) % m;
}

// ab + bq = gcd(a, b)となる(p, q)を求めて，d = gcd(a, b) をリターンする
long long extgcd(long long a, long long b, long long& x, long long& y) {
    long long d = a;
    if (b != 0) {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    } else {
        x = 1; y = 0;
    }
    return d;
}

// 中国剰余定理
// リターン値を(r, m)とすると解は x ≡ r (mod. m)
// 解ナシの場合は(0, -1)をリターン
// 2元
pair<long long, long long> chineserem(long long b1, long long m1, long long b2, long long m2) {
    long long p, q;
    long long d = extgcd(m1, m2, p, q);
    if ((b2 - b1)%d != 0) return make_pair(0, -1);
    long long m = m1 * (m2/d);
    long long tmp = (b2 - b1) / d * p % (m2/d);
    long long r = mod(b1 + m1 * tmp, m);
    return make_pair(r, m);
}

pair<long long, long long> chineserem(const vector<long long> &b, const vector<long long> &m) {
    long long r = 0, M = 1;
    for (int i = 0; i < b.size(); ++i) {
        long long p, q;
        long long d = extgcd(M, m[i], p, q);
        if ((b[i] - r) % d != 0) return make_pair(0, -1);
        long long tmp = (b[i] - r) / d * p % (m[i]/d);
        r += M * tmp;
        M *= m[i]/d;
    }
    return make_pair(mod(r, M), M);
}

// 逆元
// (a, m)が互いに素
long long modinv(long long a, long long m) {
    long long x, y;
    extgcd(a, m, x, y);
    return mod(x, m);
}

// mは互いに素でなければいけない
long long garner(vector<long long> b, vector<long long> m, long long MOD) {
    m.push_back(MOD);
    vector<long long> coeffs((int)m.size(), 1);
    vector<long long> constants((int)m.size(), 0);
    for (int k = 0; k < b.size(); ++k) {
        long long t = mod((b[k] - constants[k]) * modinv(coeffs[k], m[k]), m[k]);
        for (int i = k+1; i<m.size(); ++i) {
            (constants[i] += t * coeffs[i]) %= m[i];
            (coeffs[i] *= m[k]) %= m[i];
        }
    }
    return constants.back();
}


long long gcd(long long a, long long b) {
    if (b == 0) return a;
    else return gcd(b, a%b);
}

// m[i]のlcmが帰る（？）
long long pregarner(vector<long long> &b, vector<long long> &m, long long MOD) {
    long long res = 1;
    for (int i = 0; i < b.size(); ++i) {
        for (int j = 0; j < i; ++j) {
            long long g = gcd(m[i], m[j]);

            if ((b[i] - b[j])% g != 0) return -1;

            m[i] /= g;
            m[j] /= g;

            long long gi = gcd(m[i], g), gj = g/gi;

            do {
                g = gcd(gi, gj);
                gi *= g;
                gj /= g;
            } while (g != 1);

            m[i] *= gi;
            m[j] *= gj;

            b[i] %= m[i];
            b[j] %= m[j];
        }
    }
    for (int i = 0; i < b.size(); ++i) (res *= m[i]) %= MOD;
    return res;
}

int main() {
    int N;
    cin >> N;
    vector<ll> x(N), y(N);
    bool nonzero = false;
    rep(i, N) {
        cin >> x[i] >> y[i];
        if (x[i] > 0) nonzero = true;
    }

    ll lcm = pregarner(x, y, 1e9+7);

    if (!nonzero) cout << lcm << endl;
    else if (lcm == -1) cout << -1 << endl;
    else cout << garner(x, y, 1e9+7) << endl;
}