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 NN = 100 + 117;
const int MM = 3000 + 117;
int read() {
  int fl = 1, x;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  if (c == '-') {
    fl = -1;
    c = getchar();
  }
  for (x = 0; c >= '0' && c <= '9'; c = getchar())
    x = (x << 3) + (x << 1) + c - '0';
  return x * fl;
}
void open() { freopen("a.in", "r", stdin); }
void close() {
  fclose(stdin);
  fclose(stdout);
}
const int mod = 1e9 + 7;
long long ksm(long long a, long long b) {
  long long ret = 1;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) ret = ret * a % mod;
  return ret;
}
int m, n;
struct eq {
  long long st, d;
} a[NN][MM];
int d[MM] = {};
int dcnt = 0;
long long in[NN][2] = {};
long long ans[MM] = {};
bool vis[MM] = {};
void dvd(int a) {
  int top = sqrt(a + 1);
  for (int i = 2; i <= top; ++i) {
    if (a % i == 0) {
      d[++dcnt] = i;
      while (a % i == 0) a /= i;
    }
  }
  if (a != 1) d[++dcnt] = a;
}
bool chk() {
  for (int i = 1; i <= n; ++i) {
    long long x = -1;
    for (int j = 1; j <= dcnt; ++j) {
      if (ans[j] < a[i][j].st) return false;
      if (a[i][j].d) {
        if ((ans[j] - a[i][j].st) % a[i][j].d) return false;
        long long now = (ans[j] - a[i][j].st) / a[i][j].d;
        if (x == -1)
          now = x;
        else if (now != x)
          return false;
      } else {
        if (ans[j] != a[i][j].st) return false;
      }
    }
  }
  return true;
}
void makeans(int id, long long x) {
  for (int i = 1; i <= dcnt; ++i) {
    ans[i] = a[id][i].st + a[id][i].d * x;
  }
}
void WA() {
  printf("-1\n");
  exit(0);
}
void AC() {
  long long op = 1;
  for (int i = 1; i <= dcnt; ++i) {
    op = op * ksm(d[i], ans[i]) % mod;
  }
  printf("%lld\n", op);
  exit(0);
}
void judge() {
  if (chk())
    AC();
  else
    WA();
}
long long x[2][3] = {};
int xiao() {
  long long mul1 = x[1][0];
  long long mul2 = x[0][0];
  x[0][0] *= mul1, x[0][1] *= mul1, x[0][2] *= mul1;
  x[1][0] *= mul2, x[1][1] *= mul2, x[1][2] *= mul2;
  x[1][0] -= x[0][0], x[1][1] -= x[0][1], x[1][2] -= x[0][2];
  if (x[1][1] == 0) {
    if (x[1][2] == 0)
      return 1;
    else
      return -1;
  }
  if (x[1][2] % x[1][1]) return -1;
  x[1][2] /= x[1][1];
  if (x[1][2] < 0) return -1;
  x[0][2] -= x[0][1] * x[1][2];
  if (x[0][2] % x[0][0]) return -1;
  x[0][2] /= x[0][0];
  if (x[0][2] < 0) return -1;
  return 0;
}
long long exgcd(long long a, long long &x, long long b, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long nx, ny;
  long long g = exgcd(b, nx, a % b, ny);
  x = ny;
  y = nx - a / b * ny;
  return g;
}
long long mul(long long a, long long b, long long mod) {
  long long ret = 0;
  if (b < 0) b = -b, a = -a;
  for (; b; b >>= 1, a = (a + a) % mod) {
    if (b & 1) ret = (ret + a) % mod;
  }
  return ret;
}
bool bind(eq &a, eq b) {
  long long x, y;
  long long g = exgcd(a.d, x, b.d, y);
  long long c = b.st - a.st;
  if (c % g) return false;
  c /= g;
  long long mod = a.d / g * b.d;
  long long tp = mul(a.d, x, mod);
  tp = mul(tp, c, mod);
  tp = (tp + a.st) % mod;
  long long rem = tp;
  a.st = rem;
  a.d = mod;
  return true;
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    int a = read();
    int b = read();
    dvd(a);
    dvd(b);
    in[i][0] = a;
    in[i][1] = b;
  }
  sort(d + 1, d + 1 + dcnt);
  dcnt = unique(d + 1, d + 1 + dcnt) - d - 1;
  for (int i = 1; i <= n; ++i) {
    long long x = in[i][0], y = in[i][1];
    for (int j = 1; j <= dcnt; ++j) {
      while (x % d[j] == 0) {
        a[i][j].st++;
        x /= d[j];
      }
      while (y % d[j] == 0) {
        a[i][j].d++;
        y /= d[j];
      }
    }
  }
  if (n == 1) {
    makeans(1, 0);
    judge();
  }
  int rem = dcnt;
  for (int j = 1; j <= dcnt; ++j) {
    int k = -1;
    int pos = -1;
    for (int i = 1; i <= n; ++i) {
      if (a[i][j].d == 0) {
        if (k == -1) {
          k = a[i][j].st;
        } else {
          if (a[i][j].st != k) WA();
        }
      } else
        pos = i;
    }
    if (k == -1) continue;
    if (pos == -1) {
      vis[j] = 1;
      --rem;
      continue;
    }
    if (k < a[pos][j].st) {
      WA();
    }
    k -= a[pos][j].st;
    if (k % a[pos][j].d) WA();
    k /= a[pos][j].d;
    makeans(pos, k);
    judge();
  }
  if (rem == 0) {
    makeans(1, 0);
    judge();
  }
  if (rem > 1) {
    int pos = 0;
    for (int j = 1; j <= dcnt; ++j)
      if (!vis[j]) {
        pos = j;
        break;
      }
    for (int j = pos + 1; j <= dcnt; ++j) {
      if (vis[j]) continue;
      for (int i = 2; i <= n; ++i) {
        x[0][0] = a[1][pos].d;
        x[0][1] = -a[i][pos].d;
        x[0][2] = a[i][pos].st - a[1][pos].st;
        x[1][0] = a[1][j].d;
        x[1][1] = -a[i][j].d;
        x[1][2] = a[i][j].st - a[1][j].st;
        int tp = xiao();
        if (tp == 1) vis[j] = 1;
        if (tp == -1) WA();
        if (tp == 0) {
          makeans(1, x[0][2]);
          judge();
        }
      }
    }
  }
  int pos = 0;
  for (int j = 1; j <= dcnt; ++j)
    if (!vis[j]) {
      pos = j;
      break;
    }
  eq now;
  now.st = 0;
  now.d = 1;
  for (int i = 1; i <= n; ++i) {
    if (!bind(now, a[i][pos])) WA();
  }
  for (int i = 1; i <= n; ++i) {
    while (now.st < a[i][pos].st) {
      now.st += now.d;
    }
  }
  long long x = (now.st - a[1][pos].st) / a[1][pos].d;
  makeans(1, x);
  judge();
  close();
  return 0;
}
