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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long exgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long g = exgcd(b, a % b, y, x);
  y -= a / b * x;
  return g;
}
void gg() {
  printf("-1\n");
  exit(0);
}
int p[6010], N;
long long A[6010], B[6010];
struct Eq {
  long long a, b, c;
} e[6010];
long long chu(long long a, long long b) { return (a + b - 1) / b; }
long long mmul(long long a, long long b, long long M) {
  a = (a % M + M) % M;
  b = (b % M + M) % M;
  long long ans = 0, t = a;
  while (b) {
    if (b & 1) ans = (ans + t) % M;
    t = (t + t) % M;
    b >>= 1;
  }
  return ans;
}
bool merge(int a, int b) {
  int cnt = 0;
  for (int i = 1; i <= N; i++) {
    int cnt1 = 0, cnt2 = 0;
    int t = a;
    while (t % p[i] == 0) t /= p[i], cnt1++;
    t = b;
    while (t % p[i] == 0) t /= p[i], cnt2++;
    if (!B[i] && !cnt2) {
      if (cnt1 != A[i]) gg();
      continue;
    }
    cnt++;
    e[cnt].a = B[i];
    e[cnt].b = -cnt2;
    e[cnt].c = cnt1 - A[i];
  }
  if (!cnt) return false;
  bool flag = false;
  for (int i = 1; i <= cnt && !flag; i++) {
    for (int j = i + 1; j <= cnt && !flag; j++)
      if (e[i].a * e[j].b != e[j].a * e[i].b) {
        swap(e[i], e[1]);
        swap(e[j], e[2]);
        flag = true;
      } else if (e[i].a * e[j].c != e[j].a * e[i].c ||
                 e[i].b * e[j].c != e[j].b * e[i].c)
        gg();
  }
  if (flag) {
    if (!e[1].a) swap(e[1], e[2]);
    long long g = gcd(e[1].a, e[2].a);
    long long m1 = e[2].a / g;
    long long m2 = e[1].a / g;
    long long dc = e[1].c * m1 - e[2].c * m2;
    long long db = e[1].b * m1 - e[2].b * m2;
    if (dc % db != 0) gg();
    long long y = dc / db;
    dc = e[1].c - e[1].b * y;
    if (dc % e[1].a != 0) gg();
    long long x = dc / e[1].a;
    for (int i = 1; i <= cnt; i++)
      if (x * e[i].a + y * e[i].b != e[i].c) gg();
    if (x < 0 || y < 0) gg();
    for (int i = 1; i <= N; i++) A[i] += B[i] * x;
    return true;
  } else {
    long long a = e[1].a, b = -e[1].b, c = e[1].c;
    if (a == 0) {
      if (c % b != 0) gg();
      if (c / b < 0) gg();
      for (int i = 1; i <= N; i++) {
        int cnt1 = 0, cnt2 = 0;
        int t = a;
        while (t % p[i] == 0) t /= p[i], cnt1++;
        t = b;
        while (t % p[i] == 0) t /= p[i], cnt2++;
        A[i] = cnt1 + cnt2 * (c / b);
      }
      return true;
    }
    if (b == 0) {
      if (c % a != 0) gg();
      if (c / a < 0) gg();
      for (int i = 1; i <= N; i++) A[i] += c / a * B[i];
      return true;
    }
    long long x, y;
    long long g = exgcd(a, b, x, y);
    if (c % g != 0) gg();
    x = mmul(x, c / g, b / g);
    y = -(c - x * a) / b;
    long long d = chu(max(0ll, -y), a / g);
    x += d * (b / g);
    for (int i = 1; i <= N; i++) {
      A[i] += B[i] * x;
      B[i] *= b / g;
    }
    return false;
  }
}
const int Mod = 1000000007;
int fpow(int a, long long b) {
  int ans = 1, t = a;
  while (b) {
    if (b & 1) ans = (long long)ans * t % Mod;
    t = (long long)t * t % Mod;
    b >>= 1;
  }
  return ans;
}
void work(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) {
      p[++N] = i;
      while (x % i == 0) x /= i;
    }
  if (x > 1) p[++N] = x;
  return;
}
int a[110], b[110];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &a[i], &b[i]);
    work(a[i]);
    work(b[i]);
  }
  sort(p + 1, p + N + 1);
  N = unique(p + 1, p + N + 1) - p - 1;
  for (int i = 1; i <= N; i++) {
    int t = a[1];
    while (t % p[i] == 0) A[i]++, t /= p[i];
    t = b[1];
    while (t % p[i] == 0) B[i]++, t /= p[i];
  }
  int now = 2;
  while (now <= n && !merge(a[now], b[now])) now++;
  while (now <= n) {
    for (int i = 1; i <= N; i++) {
      int cnt1 = 0, cnt2 = 0;
      int t = a[now];
      while (t % p[i] == 0) t /= p[i], cnt1++;
      t = b[now];
      while (t % p[i] == 0) t /= p[i], cnt2++;
      if (!cnt2) {
        if (cnt1 != A[i]) gg();
      } else if (A[i] < cnt1 || A[i] % cnt2 != cnt1 % cnt2)
        gg();
    }
    now++;
  }
  int ans = 1;
  for (int i = 1; i <= N; i++) ans = (long long)ans * fpow(p[i], A[i]) % Mod;
  printf("%d\n", ans);
  return 0;
}
