#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int n;
set<int> all;
struct d {
  map<int, int> v;
  int& operator[](int x) { return v[x]; }
  void init(int x) {
    for (int i = 2; i * i <= x; i++)
      if (x % i == 0) {
        int cnt = 0;
        while (x % i == 0) cnt++, x /= i;
        v[i] = cnt;
        all.insert(i);
      }
    if (x > 1) {
      v[x] = 1;
      all.insert(x);
    }
  }
  bool operator!=(d b) {
    for (set<int>::iterator it = all.begin(); it != all.end(); it++)
      if (v[*it] != b[*it]) return 1;
    return 0;
  }
};
long long gcd(long long a, long long b) { return (b ? gcd(b, a % b) : a); }
int trs(d& a) {
  int g = 0;
  for (map<int, int>::iterator it = a.v.begin(); it != a.v.end(); it++)
    g = gcd(g, it->second);
  if (g)
    for (map<int, int>::iterator it = a.v.begin(); it != a.v.end(); it++)
      it->second /= g;
  return g;
}
d a[105], b[105];
inline int power(int x, long long y) {
  int z = 1;
  while (y) {
    if (y & 1) z = 1ll * z * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return z;
}
namespace unsame {
void chk(d a0, d a1, d b0, d b1) {
  int aa, bb, cc;
  bool ok = 0;
  int x, y;
  for (set<int>::iterator it = all.begin(); it != all.end(); it++) {
    int ta = b0[*it];
    int tb = -b1[*it];
    int tc = a0[*it] - a1[*it];
    if (ta || tb || tc) {
      if (!ta || !tb) {
        if (!ta && !tb) {
          puts("-1");
          return;
        }
        if (!ta) {
          x = -tc / tb;
          if (x < 0 || tc % tb) {
            puts("-1");
            return;
          }
          swap(a0, b0);
        } else {
          x = -tc / ta;
          if (x < 0 || tc % ta) {
            puts("-1");
            return;
          }
        }
        break;
      }
      int g = gcd(gcd(ta, tb), tc);
      ta /= g;
      tb /= g;
      tc /= g;
      if (ok && (ta != aa || tb != bb || tc != cc)) {
        int t = ta * bb - tb * aa;
        if (!t) {
          puts("-1");
          return;
        }
        int w = aa * tc - ta * cc;
        y = w / t;
        if (y < 0 || y * t != w) {
          puts("-1");
          return;
        }
        w = tb * cc - bb * tc;
        x = w / t;
        if (x < 0 || x * t != w) {
          puts("-1");
          return;
        }
        break;
      } else {
        aa = ta;
        bb = tb;
        cc = tc;
        ok = 1;
      }
    }
  }
  d r;
  for (set<int>::iterator it = all.begin(); it != all.end(); it++)
    r[*it] = a0[*it] + x * b0[*it];
  for (int i = 1; i <= n; i++) {
    int p = -1;
    for (set<int>::iterator it = all.begin(); it != all.end(); it++) {
      if (b[i][*it]) {
        int t = (r[*it] - a[i][*it]) / b[i][*it];
        if (t * b[i][*it] + a[i][*it] != r[*it]) {
          puts("-1");
          return;
        }
        if (p == -1) p = t;
        if (t != p) {
          puts("-1");
          return;
        }
      } else {
        if (r[*it] != a[i][*it]) {
          puts("-1");
          return;
        }
      }
    }
  }
  int res = 1;
  for (set<int>::iterator it = all.begin(); it != all.end(); it++) {
    res = 1ll * res * power(*it, r[*it]) % mod;
  }
  printf("%d\n", res);
}
}  // namespace unsame
namespace same {
int k[105], c[105];
int T(d& a, d& b) {
  int mn = 1e9;
  for (set<int>::iterator it = all.begin(); it != all.end(); it++)
    if (b[*it]) {
      mn = min(mn, a[*it] / b[*it]);
    }
  for (set<int>::iterator it = all.begin(); it != all.end(); it++)
    if (b[*it]) {
      a[*it] -= mn * b[*it];
    }
  return mn;
}
void exgcd(long long a, long long b, long long& x, long long& y, long long& c) {
  if (!b) {
    y = 0;
    x = 1;
    c = a;
    return;
  }
  exgcd(b, a % b, y, x, c);
  y -= a / b * x;
}
void work() {
  for (int i = 1; i <= n; i++) c[i] = T(a[i], b[i]);
  for (int i = 2; i <= n; i++)
    if (a[i] != a[1]) {
      puts("-1");
      exit(0);
    }
  long long tk = k[1], tc = c[1];
  for (int i = 2; i <= n; i++) {
    long long g, x, y;
    exgcd(tk, -k[i], x, y, g);
    if ((c[i] - tc) % g) {
      puts("-1");
      return;
    }
    x *= (c[i] - tc) / g;
    y *= (c[i] - tc) / g;
    g = abs(g);
    k[i] /= g;
    tk /= g;
    long long t = max((-x + k[i] - 1) / k[i], (-y + tk - 1) / tk);
    x += t * k[i], y += t * tk;
    t = min(x / k[i], y / tk);
    x -= t * k[i], y -= t * tk;
    tc = tk * g * x + tc;
    tk = tk * k[i] * g;
  }
  int res = 1;
  for (map<int, int>::iterator it = a[1].v.begin(); it != a[1].v.end(); it++)
    if (it->second) {
      res = 1ll * res * power(it->first, it->second) % mod;
    }
  for (map<int, int>::iterator it = b[1].v.begin(); it != b[1].v.end(); it++)
    if (it->second) {
      res = 1ll * res * power(it->first, it->second * tc) % mod;
    }
  printf("%d\n", res);
}
}  // namespace same
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[i].init(x);
    b[i].init(y);
  }
  for (int i = 1; i <= n; i++) same::k[i] = trs(b[i]);
  for (int i = 2; i <= n; i++) {
    if (b[1] != b[i]) {
      for (int i = 1; i <= n; i++)
        for (map<int, int>::iterator it = b[i].v.begin(); it != b[i].v.end();
             it++)
          it->second *= same::k[i];
      unsame::chk(a[1], a[i], b[1], b[i]);
      return 0;
    }
  }
  same::work();
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int n;
set<int> all;
struct d {
  map<int, int> v;
  int& operator[](int x) { return v[x]; }
  void init(int x) {
    for (int i = 2; i * i <= x; i++)
      if (x % i == 0) {
        int cnt = 0;
        while (x % i == 0) cnt++, x /= i;
        v[i] = cnt;
        all.insert(i);
      }
    if (x > 1) {
      v[x] = 1;
      all.insert(x);
    }
  }
  bool operator!=(d b) {
    for (set<int>::iterator it = all.begin(); it != all.end(); it++)
      if (v[*it] != b[*it]) return 1;
    return 0;
  }
};
long long gcd(long long a, long long b) { return (b ? gcd(b, a % b) : a); }
int trs(d& a) {
  int g = 0;
  for (map<int, int>::iterator it = a.v.begin(); it != a.v.end(); it++)
    g = gcd(g, it->second);
  if (g)
    for (map<int, int>::iterator it = a.v.begin(); it != a.v.end(); it++)
      it->second /= g;
  return g;
}
d a[105], b[105];
inline int power(int x, long long y) {
  int z = 1;
  while (y) {
    if (y & 1) z = 1ll * z * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return z;
}
namespace unsame {
void chk(d a0, d a1, d b0, d b1) {
  int aa, bb, cc;
  bool ok = 0;
  int x, y;
  for (set<int>::iterator it = all.begin(); it != all.end(); it++) {
    int ta = b0[*it];
    int tb = -b1[*it];
    int tc = a0[*it] - a1[*it];
    if (ta || tb || tc) {
      if (!ta || !tb) {
        if (!ta && !tb) {
          puts("-1");
          return;
        }
        if (!ta) {
          x = -tc / tb;
          if (x < 0 || tc % tb) {
            puts("-1");
            return;
          }
          swap(a0, b0);
        } else {
          x = -tc / ta;
          if (x < 0 || tc % ta) {
            puts("-1");
            return;
          }
        }
        break;
      }
      int g = gcd(gcd(ta, tb), tc);
      ta /= g;
      tb /= g;
      tc /= g;
      if (ok && (ta != aa || tb != bb || tc != cc)) {
        int t = ta * bb - tb * aa;
        if (!t) {
          puts("-1");
          return;
        }
        int w = aa * tc - ta * cc;
        y = w / t;
        if (y < 0 || y * t != w) {
          puts("-1");
          return;
        }
        w = tb * cc - bb * tc;
        x = w / t;
        if (x < 0 || x * t != w) {
          puts("-1");
          return;
        }
        break;
      } else {
        aa = ta;
        bb = tb;
        cc = tc;
        ok = 1;
      }
    }
  }
  d r;
  for (set<int>::iterator it = all.begin(); it != all.end(); it++)
    r[*it] = a0[*it] + x * b0[*it];
  for (int i = 1; i <= n; i++) {
    int p = -1;
    for (set<int>::iterator it = all.begin(); it != all.end(); it++) {
      if (b[i][*it]) {
        int t = (r[*it] - a[i][*it]) / b[i][*it];
        if (t * b[i][*it] + a[i][*it] != r[*it]) {
          puts("-1");
          return;
        }
        if (p == -1) p = t;
        if (t != p) {
          puts("-1");
          return;
        }
      } else {
        if (r[*it] != a[i][*it]) {
          puts("-1");
          return;
        }
      }
    }
  }
  int res = 1;
  for (set<int>::iterator it = all.begin(); it != all.end(); it++) {
    res = 1ll * res * power(*it, r[*it]) % mod;
  }
  printf("%d\n", res);
}
}  // namespace unsame
namespace same {
int k[105], c[105];
int T(d& a, d& b) {
  int mn = 1e9;
  for (set<int>::iterator it = all.begin(); it != all.end(); it++)
    if (b[*it]) {
      mn = min(mn, a[*it] / b[*it]);
    }
  for (set<int>::iterator it = all.begin(); it != all.end(); it++)
    if (b[*it]) {
      a[*it] -= mn * b[*it];
    }
  return mn;
}
void exgcd(long long a, long long b, long long& x, long long& y, long long& c) {
  if (!b) {
    y = 0;
    x = 1;
    c = a;
    return;
  }
  exgcd(b, a % b, y, x, c);
  y -= a / b * x;
}
void work() {
  for (int i = 1; i <= n; i++) c[i] = T(a[i], b[i]);
  for (int i = 2; i <= n; i++)
    if (a[i] != a[1]) {
      puts("-1");
      exit(0);
    }
  long long tk = k[1], tc = c[1];
  for (int i = 2; i <= n; i++) {
    long long g, x, y;
    exgcd(tk, -k[i], x, y, g);
    if ((c[i] - tc) % g) {
      puts("-1");
      return;
    }
    x *= (c[i] - tc) / g;
    y *= (c[i] - tc) / g;
    g = abs(g);
    k[i] /= g;
    tk /= g;
    long long t = max((-x + k[i] - 1) / k[i], (-y + tk - 1) / tk);
    x += t * k[i], y += t * tk;
    t = min(x / k[i], y / tk);
    x -= t * k[i], y -= t * tk;
    tc = tk * g * x + tc;
    tk = tk * k[i] * g;
  }
  int res = 1;
  for (map<int, int>::iterator it = a[1].v.begin(); it != a[1].v.end(); it++)
    if (it->second) {
      res = 1ll * res * power(it->first, it->second) % mod;
    }
  for (map<int, int>::iterator it = b[1].v.begin(); it != b[1].v.end(); it++)
    if (it->second) {
      res = 1ll * res * power(it->first, it->second * tc) % mod;
    }
  printf("%d\n", res);
}
}  // namespace same
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[i].init(x);
    b[i].init(y);
  }
  for (int i = 1; i <= n; i++) same::k[i] = trs(b[i]);
  for (int i = 2; i <= n; i++) {
    if (b[1] != b[i]) {
      for (int i = 1; i <= n; i++)
        for (map<int, int>::iterator it = b[i].v.begin(); it != b[i].v.end();
             it++)
          it->second *= same::k[i];
      unsame::chk(a[1], a[i], b[1], b[i]);
      return 0;
    }
  }
  same::work();
  return 0;
}

```