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 N = 105, L = 1e3 + 5;
const int mod = 1e9 + 7;
int n, va[N], vb[N], cta[N][L], ctb[N][L];
namespace utils {
template <class T>
inline void apn(T &x, const T y) {
  x = x < y ? x : y;
}
template <class T>
inline void apx(T &x, const T y) {
  x = x > y ? x : y;
}
inline int nxi() {
  int x = 0;
  char c;
  while (((c = getchar()) > '9' || c < '0') && c != '-')
    ;
  const bool f = c == '-' && (c = getchar());
  while (x = x * 10 - 48 + c, (c = getchar()) >= '0' && c <= '9')
    ;
  return f ? -x : x;
}
}  // namespace utils
using namespace utils;
namespace D {
int cnt, v[L];
inline void add(const int x) { v[++cnt] = x; }
inline void build() {
  sort(v + 1, v + cnt + 1);
  cnt = unique(v + 1, v + cnt + 1) - v - 1;
}
inline int ask(const int x) { return lower_bound(v + 1, v + cnt + 1, x) - v; }
}  // namespace D
void getzyz(int x) {
  for (int i = 2; i <= 3.5e4; ++i) {
    if (x % i) continue;
    D::add(i);
    while (x % i == 0) x /= i;
  }
  if (x > 1) D::add(x);
}
int qcnt(int t, const int z) {
  int ans = 0;
  assert(t);
  for (; t % z == 0; t /= z, ++ans)
    ;
  return ans;
}
inline int fpow(int x, long long t) {
  int ans = 1;
  for (; t; x = (long long)x * x % mod, t >>= 1) {
    if (t & 1) ans = (long long)ans * x % mod;
  }
  return ans;
}
inline long long smul(long long x, long long y, const long long mod) {
  long long ans = 0;
  bool f = (y < 0) ^ (x < 0);
  if (y < 0) y = -y;
  for (; y; x = x * 4 % mod, y >>= 2) {
    ans = (ans + x * (y & 3)) % mod;
  }
  ans = f ? mod - ans : ans;
  return ans;
}
int qans(const int p, const long long tot) {
  static long long tgt[L];
  int ans = va[p];
  for (int i = 1; i <= D::cnt; ++i) {
    tgt[i] = qcnt(va[p], D::v[i]) + qcnt(vb[p], D::v[i]) * tot;
    while (ans % D::v[i] == 0) ans /= D::v[i];
  }
  for (int i = 1; i <= n; ++i) {
    int cc = -1;
    for (int j = 1; j <= D::cnt; ++j) {
      int wa = qcnt(va[i], D::v[j]), wb = qcnt(vb[i], D::v[j]);
      if (!wb) {
        if (wa != tgt[j]) return -1;
        continue;
      }
      long long curt = tgt[j] - wa;
      if (curt < 0 || curt % wb) {
        return -1;
      }
      curt /= wb;
      if (cc == -1)
        cc = curt;
      else if (cc != curt)
        return -1;
    }
  }
  for (int i = 1; i <= D::cnt; ++i) {
    ans = (long long)ans * fpow(D::v[i], tgt[i]) % mod;
  }
  return ans;
}
bool san0() {
  for (int i = 1; i <= D::cnt; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (vb[j] % D::v[i] == 0) continue;
      int tgt = qcnt(va[j], D::v[i]), p = 1;
      for (; p <= n && vb[p] % D::v[i]; ++p)
        ;
      if (p > n) {
        for (int k = 2; k <= n; ++k) {
          if (qcnt(va[k], D::v[i]) != qcnt(va[1], D::v[i]))
            return puts("-1"), 0;
        }
        for (int k = i + 1; k <= D::cnt; ++k) {
          D::v[k - 1] = D::v[k];
        }
        --D::cnt, --i;
        break;
      }
      int tot = tgt - qcnt(va[p], D::v[i]);
      if (tot % qcnt(vb[p], D::v[i])) return puts("-1"), 0;
      tot /= qcnt(vb[p], D::v[i]);
      return printf("%d\n", qans(p, tot)), 0;
    }
  }
  return 1;
}
bool sanmulp() {
  using namespace D;
  for (int i = 2; i <= cnt; ++i) {
    for (int j = 2; j <= n; ++j) {
      if (ctb[1][i] * ctb[j][1] == ctb[1][1] * ctb[j][i]) {
        int tmp1 = cta[1][i] * ctb[1][1] - ctb[1][i] * cta[1][1];
        int tmp2 = cta[j][i] * ctb[1][1] - ctb[1][i] * cta[j][1];
        if (tmp1 != tmp2) return puts("-1"), 0;
      } else {
        int numj = cta[1][i] * ctb[1][1] - cta[1][1] * ctb[1][i];
        numj -= cta[j][i] * ctb[1][1] - cta[j][1] * ctb[1][i];
        int denomj = ctb[j][i] * ctb[1][1] - ctb[j][1] * ctb[1][i];
        if (numj % denomj || numj / denomj < 0) return puts("-1"), 0;
        return printf("%d\n", qans(j, numj / denomj)), 0;
      }
    }
  }
  return 1;
}
long long exgcd(long long x, long long y, long long &a, long long &b) {
  if (!y) {
    a = 1, b = 0;
    return x;
  }
  long long res = exgcd(y, x % y, a, b), tmp = a - (x / y) * b;
  a = b, b = tmp;
  return res;
}
bool merge(long long &m1, long long &r1, long long m2, long long r2) {
  long long a, b, g = exgcd(m1, -m2, a, b);
  if ((r2 - r1) % g) return 0;
  m1 = m1 / g * m2;
  if (m1 < 0) m1 = -m1;
  a = smul(a, (r2 - r1) / g, m1), b = smul(b, (r2 - r1) / g, m1);
  r1 = ((smul(b, m2, m1) + r2) % m1 + m1) % m1;
  return 1;
}
int main() {
  n = nxi();
  for (int i = 1; i <= n; ++i) {
    va[i] = nxi(), vb[i] = nxi();
    getzyz(va[i]), getzyz(vb[i]);
  }
  D::build();
  if (!san0()) return 0;
  if (!D::cnt) return printf("%d\n", qans(1, 0)), 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= D::cnt; ++j) {
      cta[i][j] = qcnt(va[i], D::v[j]);
      ctb[i][j] = qcnt(vb[i], D::v[j]);
    }
  }
  if (!sanmulp()) return 0;
  {
    long long st = 1, rs = 0;
    for (int i = 1; i <= n; ++i) {
      int si = ctb[i][1], ri = cta[i][1] % si;
      if (!merge(st, rs, si, ri)) return puts("-1"), 0;
    }
    for (int i = 1; i <= n; ++i) {
      while (rs < cta[i][1]) rs += st;
    }
    for (int i = 1; i <= n; ++i) {
      assert((rs - cta[i][1]) % ctb[i][1] == 0);
    }
    printf("%d\n", qans(1, (rs - cta[1][1]) / ctb[1][1]));
  }
  return 0;
}
