Geometric progression with the first element a and common ratio b is a sequence of numbers a, ab, ab2, ab3, ....

You are given n integer geometric progressions. Your task is to find the smallest integer x, that is the element of all the given progressions, or else state that such integer does not exist.

Input

The first line contains integer (1 ≤ n ≤ 100) — the number of geometric progressions. 

Next n lines contain pairs of integers a, b (1 ≤ a, b ≤ 109), that are the first element and the common ratio of the corresponding geometric progression.

Output

If the intersection of all progressions is empty, then print  - 1, otherwise print the remainder of the minimal positive integer number belonging to all progressions modulo 1000000007 (109 + 7).

Examples

Input

2
2 2
4 1


Output

4


Input

2
2 2
3 3


Output

-1

Note

In the second sample test one of the progressions contains only powers of two, the other one contains only powers of three.
#include <bits/stdc++.h>
using namespace std;
const int N = 2111;
const int mod = 1e9 + 7;
int n;
int pri[N], tot;
long long a[N], b[N];
struct node {
  long long k, b;
  node(long long _k = 0, long long _b = 0) {
    k = _k;
    b = _b;
  }
} f[N], g[N];
long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
long long ksm(long long base, long long x) {
  long long ans = 1;
  while (x) {
    if (x & 1) ans = ans * base % mod;
    base = base * base % mod;
    x >>= 1;
  }
  return ans;
}
void get(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      pri[++tot] = i;
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) pri[++tot] = x;
}
void get_g(int id) {
  for (int i = 1; i <= tot; i++) {
    g[i].b = g[i].k = 0;
    while (a[id] % pri[i] == 0) {
      g[i].b++;
      a[id] /= pri[i];
    }
    while (b[id] % pri[i] == 0) {
      g[i].k++;
      b[id] /= pri[i];
    }
  }
}
long long ex_gcd(long long A, long long B, long long &x, long long &y) {
  if (!B) {
    x = 1;
    y = 0;
    return A;
  }
  long long xx, yy;
  long long d = ex_gcd(B, A % B, xx, yy);
  x = yy;
  y = xx - A / B * yy;
  return d;
}
bool merge() {
  bool flag = 0;
  for (int t = 0; t < 3; t++) {
    for (int i = 1; i <= tot; i++) {
      long long fk = f[i].k, fb = f[i].b;
      long long gk = g[i].k, gb = g[i].b;
      bool fl = 0;
      if (!fk && !gk) {
        if (fb != gb) return 0;
      } else {
        if (!gk || !fk) {
          if (!fk) swap(fk, gk), swap(fb, gb), fl = 1;
          if (fb > gb || (gb - fb) % fk != 0) return 0;
          long long k = (gb - fb) / fk;
          if (fl)
            for (int j = 1; j <= tot; j++) g[j].b += g[j].k * k, g[j].k = 0;
          else
            for (int j = 1; j <= tot; j++) f[j].b += f[j].k * k, f[j].k = 0;
          flag = 1;
        }
      }
    }
  }
  if (flag) return 1;
  int fl = 0;
  long long A = 0, B = 0, C = 0;
  for (int i = 1; i <= tot; i++) {
    if (!f[i].k || !g[i].k) continue;
    if (!fl)
      A = f[i].k, B = -g[i].k, C = g[i].b - f[i].b, fl = 1;
    else if (fl == 1) {
      long long nA = f[i].k, nB = -g[i].k, nC = g[i].b - f[i].b;
      long long d = gcd(nA, A);
      A *= nA / d;
      B *= nA / d;
      C *= nA / d;
      nA *= A / d;
      nB *= A / d;
      nC *= A / d;
      if (B == nB) {
        if (C != nC) return 0;
        long long d = gcd(A, gcd(abs(B), abs(C)));
        A /= d;
        B /= d;
        C /= d;
        continue;
      } else {
        if ((C - nC) % (B - nB)) return 0;
        long long y = (C - nC) / (B - nB);
        if (y < 0) return 0;
        if ((C - B * y) % A) return 0;
        long long x = (C - B * y) / A;
        if (x < 0) return 0;
        for (int j = 1; j <= tot; j++) f[j].b += f[j].k * x, f[j].k = 0;
        for (int j = i + 1; j <= tot; j++) {
          nA = f[j].k, nB = -g[j].k, nC = g[j].b - f[j].b;
          if (nA * x + nB * y != nC) return 0;
        }
        return 1;
      }
    }
  }
  long long x = -1, y = -1;
  long long d = ex_gcd(A, -B, x, y);
  if (C % d) return 0;
  y *= -1;
  long long kx = (-B) / d;
  long long ky = A / d;
  x *= C / d, y *= C / d;
  long long tx = x < 0 ? -(-x + kx - 1) / kx : x / kx;
  long long ty = y < 0 ? -(-y + ky - 1) / ky : y / ky;
  x -= min(tx, ty) * kx;
  y -= min(tx, ty) * ky;
  for (int i = 1; i <= tot; i++) {
    f[i].b += f[i].k * x;
    f[i].k = f[i].k * kx;
  }
  return 1;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i], &b[i]);
    get(a[i]);
    get(b[i]);
  }
  sort(pri + 1, pri + 1 + tot);
  tot = unique(pri + 1, pri + 1 + tot) - (pri + 1);
  for (int i = 1; i <= tot; i++) {
    while (a[1] % pri[i] == 0) {
      f[i].b++;
      a[1] /= pri[i];
    }
    while (b[1] % pri[i] == 0) {
      f[i].k++;
      b[1] /= pri[i];
    }
  }
  for (int i = 2; i <= n; i++) {
    get_g(i);
    if (!merge()) {
      puts("-1");
      return 0;
    }
  }
  int ans = 1;
  for (int i = 1; i <= tot; i++) {
    ans = 1LL * ans * ksm(pri[i], f[i].b) % mod;
  }
  cout << ans << endl;
  return 0;
}
