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;
void GG() {
  puts("-1");
  exit(0);
}
long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
void exgcd(long long x, long long y, long long &a, long long &b) {
  if (!y) {
    a = 1;
    b = 0;
    return;
  }
  exgcd(y, x % y, b, a);
  b -= (x / y) * a;
}
long long getinv(long long x, long long y) {
  long long a, b;
  exgcd(x, y, a, b);
  return (a % y + y) % y;
}
long long mul(long long x, long long y, long long mod) {
  long long s = 0;
  for (; y; y /= 2, x = (x + x) % mod)
    if (y & 1) s = (s + x) % mod;
  return s;
}
pair<long long, long long> operator/(pair<long long, long long> a,
                                     pair<long long, long long> b) {
  assert(a.first % b.first == 0);
  assert(a.second >= b.second);
  a.second -= b.second;
  assert(a.second % b.first == 0);
  return pair<long long, long long>(a.first / b.first, a.second / b.first);
}
pair<long long, long long> operator*(pair<long long, long long> a,
                                     pair<long long, long long> b) {
  return pair<long long, long long>(a.first * b.first,
                                    a.first * b.second + a.second);
}
void equation(pair<long long, long long> x, pair<long long, long long> y,
              pair<long long, long long> &v1, pair<long long, long long> &v2) {
  bool fl = (x < y);
  if (fl) swap(x, y);
  if (x == y) {
    if (!x.first)
      v1 = v2 = pair<long long, long long>(-1, 0);
    else
      v1 = v2 = pair<long long, long long>(1, 0);
  } else if (!x.first)
    GG();
  else if (!y.first) {
    if (x.second > y.second || (y.second - x.second) % x.first) GG();
    v1 = pair<long long, long long>(0, (y.second - x.second) / x.first);
    v2 = pair<long long, long long>(1, 0);
  } else if (x.first == y.first) {
    if ((x.second - y.second) % x.first) GG();
    v1 = pair<long long, long long>(1, 0);
    v2 = pair<long long, long long>(1, (x.second - y.second) / x.first);
  } else {
    long long G = gcd(x.first, y.first);
    if (x.second % G != y.second % G) GG();
    long long rem = x.second % G;
    pair<long long, long long> X = x, Y = y;
    x.first /= G;
    y.first /= G;
    x.second /= G;
    y.second /= G;
    long long mod = x.first * y.first;
    long long val = (mul(x.second % x.first,
                         mul(y.first, getinv(y.first, x.first), mod), mod) +
                     mul(y.second % y.first,
                         mul(x.first, getinv(x.first, y.first), mod), mod)) %
                    mod;
    if (val < x.second) val += ((x.second - val - 1) / mod + 1) * mod;
    if (val < y.second) val += ((y.second - val - 1) / mod + 1) * mod;
    pair<long long, long long> res(mod * G, val * G + rem);
    v1 = res / X;
    v2 = res / Y;
  }
  if (fl) swap(v1, v2);
}
void equationeq(pair<long long, long long> x, pair<long long, long long> y,
                pair<long long, long long> &v1,
                pair<long long, long long> &v2) {
  if (x == y)
    v1 = v2 = pair<long long, long long>(1, 0);
  else {
    if (x.first == y.first) GG();
    long long T = (x.second - y.second) / (y.first - x.first);
    if (x.first * T + x.second != y.first * T + y.second || T < 0) GG();
    v1 = v2 = pair<long long, long long>(0, T);
  }
}
void merge(pair<long long, long long> &v1, pair<long long, long long> &v2,
           pair<long long, long long> l1, pair<long long, long long> l2) {
  if (v1.first == -1) {
    v1 = l1;
    v2 = l2;
    return;
  }
  if (l1.first == -1) return;
  pair<long long, long long> i1, i2;
  equation(v1, l1, i1, i2);
  if (i1.first != -1) v1 = v1 * i1, l1 = l1 * i2, v2 = v2 * i1, l2 = l2 * i2;
  equationeq(v2, l2, i1, i2);
  v1 = v1 * i1;
  l1 = l1 * i2;
  v2 = v2 * i1;
  l2 = l2 * i2;
}
map<int, pair<long long, long long> > tmp, ans;
void input(map<int, pair<long long, long long> > &mp) {
  int x;
  mp.clear();
  scanf("%d", &x);
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) {
      int cnt = 0;
      for (; x % i == 0; x /= i, ++cnt)
        ;
      mp[i].second = cnt;
    }
  if (x != 1) mp[x].second++;
  scanf("%d", &x);
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) {
      int cnt = 0;
      for (; x % i == 0; x /= i, ++cnt)
        ;
      mp[i].first = cnt;
    }
  if (x != 1) mp[x].first++;
}
map<int, pair<long long, long long> > merge(
    map<int, pair<long long, long long> > x,
    map<int, pair<long long, long long> > y) {
  for (auto i : y) x[i.first].first += 0;
  pair<long long, long long> lim1(-1, 0), lim2(-1, 0);
  for (auto i : x) {
    pair<long long, long long> v1 = i.second, v2 = y[i.first], l1, l2;
    equation(v1, v2, l1, l2);
    merge(lim1, lim2, l1, l2);
  }
  if (lim1 != pair<long long, long long>(-1, 0)) {
    for (auto &i : x) i.second = i.second * lim1;
  }
  return x;
}
const int mo = 1000000007;
int power(int x, long long y) {
  int s = 1;
  for (; y; y /= 2, x = 1ll * x * x % mo)
    if (y & 1) s = 1ll * s * x % mo;
  return s;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (int)(1); i <= (int)(n); i++) {
    input(tmp);
    if (i != 1)
      ans = merge(ans, tmp);
    else
      ans = tmp;
  }
  int res = 1;
  for (auto i : ans) res = 1ll * res * power(i.first, i.second.second) % mo;
  printf("%d\n", res);
}
