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 mod = 1000000007;
int n;
vector<pair<int, long long> > fa[100], fb[100];
long long powmod(long long a, long long k) {
  long long ret = 1;
  while (k) {
    if (k & 1) {
      ret = ret * a % mod;
    }
    a = a * a % mod;
    k >>= 1;
  }
  return ret;
}
long long exgcd(long long a, long long b, long long& x, long long& y) {
  if (!b) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, y, x);
  y -= x * (a / b);
  return d;
}
long long find(const vector<pair<int, long long> >& f, int p) {
  int pos = lower_bound(f.begin(), f.end(), make_pair(p, -1ll)) - f.begin();
  if (pos != f.size() && f[pos].first == p) {
    return f[pos].second;
  }
  return 0;
}
vector<pair<int, long long> > factor(int x) {
  vector<pair<int, long long> > ret;
  for (int i = 2; i * i <= x; ++i) {
    long long cnt = 0;
    while (x % i == 0) {
      cnt++;
      x /= i;
    }
    if (cnt) {
      ret.push_back(make_pair(i, cnt));
    }
  }
  if (x > 1) {
    ret.push_back(make_pair(x, 1));
  }
  return ret;
}
void addfirst(vector<int>& s, const vector<pair<int, long long> >& t) {
  for (int i = 0; i < t.size(); ++i) {
    s.push_back(t[i].first);
  }
}
bool solve() {
  while (n > 1) {
    int cnt = 0;
    for (int i = 0; i < n - 1; ++i) {
      vector<int> v;
      addfirst(v, fa[i]);
      addfirst(v, fb[i]);
      addfirst(v, fa[i + 1]);
      addfirst(v, fb[i + 1]);
      sort(v.begin(), v.end());
      v.erase(unique(v.begin(), v.end()), v.end());
      long long px = -1, py, sx, sy;
      for (int j = 0; j < v.size(); ++j) {
        long long a1 = find(fa[i], v[j]), a2 = find(fa[i + 1], v[j]);
        long long b1 = find(fb[i], v[j]), b2 = find(fb[i + 1], v[j]);
        long long x, y, dx, dy;
        long long d = exgcd(b1, -b2, x, y);
        if (b1 == 0 && b2 == 0) {
          if (a1 != a2) {
            return false;
          }
          continue;
        } else if (b1 == 0 && a1 >= a2 && (a1 - a2) % b2 == 0) {
          x = 0;
          y = (a1 - a2) / b2;
          dx = 1;
          dy = 0;
        } else if (b2 == 0 && a2 >= a1 && (a2 - a1) % b1 == 0) {
          x = (a2 - a1) / b1;
          y = 0;
          dx = 0;
          dy = 1;
        } else if (d == 0 || (a2 - a1) % d != 0) {
          return false;
        } else {
          x *= (a2 - a1) / d;
          y *= (a2 - a1) / d;
          dx = b2 / abs(d);
          dy = b1 / abs(d);
          if (x > 0 && dx > 0) {
            long long t = (x - 1) / dx + 1;
            x -= t * dx;
            y -= t * dy;
          }
          if (y > 0 && dy > 0) {
            long long t = (y - 1) / dy + 1;
            x -= t * dx;
            y -= t * dy;
          }
          if (x < 0) {
            if (dx == 0) {
              return false;
            }
            long long t = (-x - 1) / dx + 1;
            x += t * dx;
            y += t * dy;
          }
          if (y < 0) {
            if (dy == 0) {
              return false;
            }
            long long t = (-y - 1) / dy + 1;
            x += t * dx;
            y += t * dy;
          }
        }
        if (px == -1) {
          px = x;
          py = y;
          sx = dx;
          sy = dy;
        } else {
          if (sx != 0 || sy != 0) {
            if (sx != dx || sy != dy) {
              long long k1 =
                  ((x - px) * dy - (y - py) * dx) / (sx * dy - sy * dx);
              long long nx = px + sx * k1, ny = py + sy * k1;
              if (nx < x || ny < y || (nx - x) * dy - dx * (ny - y) != 0) {
                return false;
              }
              px = nx;
              py = ny;
              sx = sy = 0;
            } else if ((px - x) * dy - dx * (py - y) != 0) {
              return false;
            }
          } else if (px < x || py < y || (px - x) * dy - dx * (py - y) != 0) {
            return false;
          }
        }
      }
      if (px != -1) {
        vector<pair<int, long long> > nfa, nfb;
        for (int j = 0; j < v.size(); ++j) {
          long long a1 = find(fa[i], v[j]), b1 = find(fb[i], v[j]);
          if (b1 * px + a1) {
            nfa.push_back(make_pair(v[j], b1 * px + a1));
          }
          if (b1 * sx) {
            nfb.push_back(make_pair(v[j], b1 * sx));
          }
        }
        fa[cnt] = nfa;
        fb[cnt++] = nfb;
      }
    }
    n = cnt;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    fa[i] = factor(a);
    fb[i] = factor(b);
  }
  if (solve()) {
    long long ans = 1;
    for (int i = 0; i < fa[0].size(); ++i) {
      ans = ans * powmod(fa[0][i].first, fa[0][i].second) % mod;
    }
    printf("%I64d\n", ans);
  } else {
    printf("-1\n");
  }
  return 0;
}
