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>
const int INF = 1e9;
const long long lINF = 1e18;
const double EPS = 1e-12;
using namespace std;
const int N = 111;
int n;
int a[N], b[N];
map<int, long long> pa[N], push_back[N];
const int MOD = 1e9 + 7;
void makenotcollinear(int i, int p1, int p2) {
  long long x1 = pa[0][p1], y1 = pa[0][p2];
  long long dx1 = push_back[0][p1], dy1 = push_back[0][p2];
  long long x2 = pa[i][p1], y2 = pa[i][p2];
  long long dx2 = push_back[i][p1], dy2 = push_back[i][p2];
  long long normx = -dy2, normy = dx2;
  long long want = (x2 - x1) * normx + (y2 - y1) * normy;
  long long step = dx1 * normx + dy1 * normy;
  assert(step != 0);
  long long mult = want / step;
  for (auto j : push_back[i]) {
    pa[0][j.first] += push_back[0][j.first] * mult;
  }
}
int power(int a, long long p) {
  int res = 1;
  for (; p; p >>= 1) {
    if (p & 1) {
      res = (res * 1ll * a) % MOD;
    }
    a = (a * 1ll * a) % MOD;
  }
  return res;
}
long long prod(long long a, long long b, long long curmod) {
  long long res = 0;
  bool flag = false;
  if (b < 0) {
    b = -b;
    flag = true;
  }
  for (; b; b >>= 1) {
    if (b & 1) {
      res = (res + a) % curmod;
    }
    a = a * 2 % curmod;
  }
  if (flag) {
    res = -res;
  }
  return res;
}
long long gcdExt(long long a, long long b, long long& M, long long& N) {
  if (b == 0) {
    M = 1;
    N = 0;
    return a;
  } else {
    long long k = a / b;
    long long res = gcdExt(b, a - k * b, M, N);
    swap(M, N);
    N -= k * M;
    return res;
  }
}
void solve() {
  bool ok = true;
  for (int i = 1; i < n && ok; i++) {
    vector<int> primes;
    for (auto j : pa[0]) {
      primes.push_back(j.first);
    }
    for (auto j : pa[i]) {
      primes.push_back(j.first);
    }
    for (auto j : push_back[i]) {
      primes.push_back(j.first);
    }
    long long mult = -1;
    for (auto j : primes) {
      long long want = pa[0][j];
      long long have = pa[i][j];
      if (push_back[i][j] == 0) {
        if (have != want) {
          ok = false;
          break;
        }
      } else {
        if (mult != -1) {
          if (mult * push_back[i][j] + have != want) {
            ok = false;
            break;
          }
        } else {
          mult = (want - have) / push_back[i][j];
          if (mult < 0 || mult * push_back[i][j] + have != want) {
            ok = false;
            break;
          }
        }
      }
    }
  }
  if (ok) {
    int ans = 1;
    for (auto i : pa[0]) {
      ans = ans * 1ll * power(i.first, i.second) % MOD;
    }
    printf("%d\n", ans);
  } else {
    printf("-1\n");
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", a + i, b + i);
    if (b[i] != 1) {
      swap(a[i], a[0]);
      swap(b[i], b[0]);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 2; j * j <= a[i]; j++) {
      while (a[i] % j == 0) {
        a[i] /= j;
        pa[i][j]++;
      }
    }
    if (a[i] > 1) {
      pa[i][a[i]]++;
    }
    for (int j = 2; j * j <= b[i]; j++) {
      while (b[i] % j == 0) {
        b[i] /= j;
        push_back[i][j]++;
      }
    }
    if (b[i] > 1) {
      push_back[i][b[i]]++;
    }
  }
  int p1 = 2, p1cnt = 0;
  for (auto i : push_back[0]) {
    if (i.second != 0) {
      p1 = i.first;
      p1cnt = i.second;
    }
  }
  bool collinear = true;
  for (int i = 1; i < n; i++) {
    int cur = push_back[i][p1];
    for (auto j : push_back[i]) {
      if (push_back[0][j.first] * cur != j.second * p1cnt) {
        makenotcollinear(i, p1, j.first);
        collinear = false;
      }
    }
    for (auto j : push_back[0]) {
      if (j.second * cur != push_back[i][j.first] * p1cnt) {
        makenotcollinear(i, p1, j.first);
        collinear = false;
      }
    }
  }
  if (!collinear) {
    solve();
  } else {
    long long a = pa[0][p1];
    long long b = push_back[0][p1];
    if (b == 0) {
      solve();
    } else {
      a %= b;
      bool ok = true;
      for (int i = 1; i < n; i++) {
        long long c = push_back[i][p1];
        if (c == 0) {
          if (a != pa[i][p1] % b) {
            ok = false;
            break;
          } else {
            a = pa[i][p1];
            continue;
          }
        }
        long long M, N;
        long long g = gcdExt(b, c, M, N);
        long long d = pa[i][p1];
        if (a % g != d % g) {
          ok = false;
          break;
        }
        long long lcm = b / g * c;
        long long na = prod(M, b, lcm);
        na = prod(na, (d % c - a % c) / g, lcm);
        na += a;
        na %= lcm;
        if (na < 0) {
          na += lcm;
        }
        a = na;
        b = lcm;
      }
      if (!ok) {
        printf("-1\n");
      } else {
        for (int i = 0; i < n; i++) {
          while (pa[i][p1] > a) {
            a += push_back[i][p1];
          }
        }
        long long add = (a - pa[0][p1]) / push_back[0][p1];
        for (auto& i : pa[0]) {
          i.second += push_back[0][i.first] * add;
        }
        solve();
      }
    }
  }
  return 0;
}
