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 std::cerr;
using std::endl;
const int N = 1145141, P = 1e9 + 7;
inline long long fpow(long long x, long long y) {
  long long ret = 1;
  for (; y; y >>= 1, x = x * x % P)
    if (y & 1) ret = ret * x % P;
  return ret;
}
int n, A[N], B[N];
int pr[N], tot;
long long a[N], b[N], c[N], d[N];
inline void init(int x) {
  for (int i = 2; i * i <= x; ++i) {
    if (x % i) continue;
    pr[++tot] = i;
    while (!(x % i)) x /= i;
  }
  if (x > 1) pr[++tot] = x;
}
inline void work(int x, long long f[]) {
  for (int i = 1; i <= tot; ++i) {
    int p = pr[i];
    f[i] = 0;
    while (!(x % p)) x /= p, ++f[i];
  }
}
long long gcd(long long x, long long y) { return !y ? x : gcd(y, x % y); }
inline long long lcm(long long x, long long y) {
  if (!x && !y) return 0;
  return x / gcd(x, y) * y;
}
long long exgcd(long long x, long long y, long long &a, long long &b) {
  if (!y) return a = 1, b = 0, x;
  int g = exgcd(y, x % y, b, a);
  return b -= x / y * a, g;
}
inline int merge(int kase) {
  long long sumb = 0, sumd = 0;
  for (int i = 1; i <= tot; ++i) sumb += b[i], sumd += d[i];
  if (!sumb && !sumd) {
    for (int i = 1; i <= tot; ++i)
      if (a[i] != c[i]) return 0;
    return 1;
  }
  if (!sumb || !sumd) {
    if (!sumd) {
      std::swap(sumb, sumd);
      std::swap(a, c);
      std::swap(b, d);
    }
    long long y = 0;
    for (int i = 1; i <= tot; ++i) {
      if (!d[i]) {
        if (a[i] != c[i]) return 0;
      } else {
        if (a[i] < c[i] || (a[i] - c[i]) % d[i]) return 0;
        y = (a[i] - c[i]) / d[i];
        break;
      }
    }
    for (int i = 1; i <= tot; ++i)
      if (a[i] != c[i] + y * d[i]) return 0;
    return 1;
  }
  int flag = 0;
  for (int i = 1; i <= tot; ++i)
    if (b[i] * sumd != d[i] * sumb) {
      flag = i;
      break;
    }
  if (flag) {
    int pi = flag, pj = 0;
    for (int i = 1; i <= tot; ++i)
      if (b[i] * d[pi] != d[i] * b[pi]) {
        pj = i;
        break;
      }
    long long t1 = d[pj] * (c[pi] - a[pi]) - d[pi] * (c[pj] - a[pj]),
              t2 = b[pj] * (c[pi] - a[pi]) - b[pi] * (c[pj] - a[pj]),
              t3 = b[pi] * d[pj] - b[pj] * d[pi];
    if (t3 == 0 || t1 % t3 || t2 % t3) return 0;
    long long x = t1 / t3, y = t2 / t3;
    if (x < 0 || y < 0) return 0;
    for (int i = 1; i <= tot; ++i) {
      if (a[i] + x * b[i] != c[i] + y * d[i]) return 0;
      a[i] = a[i] + x * b[i], b[i] = 0;
    }
    return 1;
  }
  long long x, y, tb = 0, td = 0, t = 0, g = 0;
  for (int i = 1; i <= tot; ++i)
    if (b[i]) {
      g = gcd(b[i], d[i]);
      tb = b[i] / g, td = d[i] / g;
      if ((a[i] - c[i]) % g) return 0;
      t = (a[i] - c[i]) / g;
      break;
    }
  for (int i = 1; i <= tot; ++i)
    if (c[i] + t * g != a[i]) return 0;
  if (t < 0) {
    std::swap(a, c);
    std::swap(b, d);
    std::swap(tb, td);
    t = -t;
  }
  exgcd(tb, td, x, y);
  x = (x * -t % td + td) % td;
  for (int i = 1; i <= tot; ++i) {
    a[i] += x * b[i];
    b[i] *= td;
  }
  return 1;
}
int main() {
  std::cin >> n;
  for (int i = 1; i <= n; ++i) {
    std::cin >> A[i] >> B[i];
    init(A[i]), init(B[i]);
  }
  std::sort(pr + 1, pr + tot + 1);
  tot = std::unique(pr + 1, pr + tot + 1) - pr - 1;
  work(A[1], a), work(B[1], b);
  for (int i = 2; i <= n; ++i) {
    work(A[i], c), work(B[i], d);
    if (!merge(i)) return puts("-1"), 0;
  }
  long long ans = 1;
  for (int i = 1; i <= tot; ++i) ans = ans * fpow(pr[i], a[i]) % P;
  std::cout << ans << std::endl;
  return 0;
}
