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 Max_N(105);
int N, _A[Max_N], _B[Max_N];
map<int, long long int> A[Max_N], B[Max_N];
long long int limit;
long long int gcd(long long int a, long long int b) {
  return b == 0LL ? a : gcd(b, a % b);
}
inline long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
void GG() {
  printf("-1");
  exit(0);
}
void calc(int n, map<int, long long int> &S) {
  for (int p = 2, e; p * p <= n; ++p)
    if (n % p == 0) {
      e = 0;
      while (n % p == 0) n /= p, ++e;
      S[p] = e;
    }
  if (n > 1) S[n] = 1;
}
set<int> S;
void merge(int a, int b) {
  S.clear();
  for (map<int, long long int>::iterator it = A[a].begin(); it != A[a].end();
       ++it)
    S.insert(it->first);
  for (map<int, long long int>::iterator it = B[a].begin(); it != B[a].end();
       ++it)
    S.insert(it->first);
  for (map<int, long long int>::iterator it = A[b].begin(); it != A[b].end();
       ++it)
    S.insert(it->first);
  for (map<int, long long int>::iterator it = B[b].begin(); it != B[b].end();
       ++it)
    S.insert(it->first);
}
namespace PrintAns {
const int MOD(1000000000 + 7);
constexpr int Mult(int a, int b) { return a * 1LL * b % MOD; }
int power(int a, long long int n) {
  int Ret(1);
  while (n) {
    if (n & 1LL) Ret = Mult(Ret, a);
    a = Mult(a, a), n >>= 1LL;
  }
  return Ret;
}
void printAns(long long int x) { printf("%d", Mult(_A[1], power(_B[1], x))); }
}  // namespace PrintAns
void print(long long int x1) {
  if (x1 < 0LL) GG();
  for (int i = 2; i <= N; ++i) {
    merge(1, i);
    set<long long int> qwq;
    for (set<int>::iterator it = S.begin(); it != S.end(); ++it) {
      int p = *it;
      long long int xi = B[1][p] * x1 + A[1][p] - A[i][p];
      if (B[i][p] == 0LL)
        if (xi == 0LL)
          continue;
        else
          GG();
      if (xi < 0LL) GG();
      if (xi % B[i][p]) GG();
      qwq.insert(xi / B[i][p]);
    }
    if (qwq.size() > 1) GG();
  }
  PrintAns::printAns(x1), exit(0);
}
constexpr long long int Abs(long long int x) { return x >= 0LL ? x : -x; }
pair<long long int, long long int> get(pair<long long int, long long int> a) {
  if (a.second < 0LL) a.second = -a.second, a.first = -a.first;
  if (a.first == 0LL)
    a.second = 1LL;
  else {
    long long int d = gcd(Abs(a.first), Abs(a.second));
    a.first /= d, a.second /= d;
  }
  return a;
}
inline pair<long long int, long long int> operator-(
    const pair<long long int, long long int> &a,
    const pair<long long int, long long int> &b) {
  long long int l = lcm(a.second, b.second);
  return get(make_pair(a.first * (l / a.second) - b.first * (l / b.second), l));
}
inline pair<long long int, long long int> operator/(
    const pair<long long int, long long int> &a,
    const pair<long long int, long long int> &b) {
  return get(make_pair(a.first * b.second, a.second * b.first));
}
void exgcd(long long int a, long long int b, long long int &x,
           long long int &y) {
  if (b == 0LL)
    x = 1LL, y = 0LL;
  else
    exgcd(b, a % b, y, x), y -= x * (a / b);
}
inline long long int inverse(long long int a, long long int MOD) {
  long long int invx, invy;
  exgcd(a, MOD, invx, invy);
  return (invx % MOD + MOD) % MOD;
}
inline long long int Add(long long int a, long long int b, long long int MOD) {
  return a + b >= MOD ? a + b - MOD : a + b;
}
inline long long int Mult(long long int a, long long int b, long long int MOD) {
  return a * b % MOD;
}
long long int A1, MOD1;
inline void merge(long long int A2, long long int MOD2) {
  long long int d = gcd(MOD1, MOD2), sub = (((A2 - A1) % MOD2 + MOD2) % MOD2);
  if (sub % d) GG();
  long long int k1, k2;
  exgcd(MOD1, MOD2, k1, k2), k1 *= sub / d;
  A1 = k1 * MOD1 + A1, MOD1 = lcm(MOD1, MOD2), A1 = (A1 % MOD1 + MOD1) % MOD1;
}
inline void insert(long long int k, long long int A2, long long int MOD2) {
  k = (k % MOD2 + MOD2) % MOD2;
  A2 = (A2 % MOD2 + MOD2) % MOD2;
  if (k == 0LL)
    if (A2 == 0LL)
      return;
    else
      GG();
  long long int d = gcd(k, MOD2);
  if (A2 % d) GG();
  k /= d, A2 /= d, MOD2 /= d;
  A2 = Mult(A2, inverse(k, MOD2), MOD2);
  merge(A2, MOD2);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i)
    scanf("%d%d", _A + i, _B + i), calc(_A[i], A[i]), calc(_B[i], B[i]);
  A1 = 0LL, MOD1 = 1LL;
  for (int i = 2; i <= N; ++i) {
    merge(1, i);
    map<pair<long long int, long long int>, pair<long long int, long long int> >
        rec;
    for (set<int>::iterator it = S.begin(); it != S.end(); ++it) {
      int p = *it;
      if (B[i][p] == 0LL)
        if (B[1][p] == 0LL) {
          if (A[1][p] != A[i][p]) GG();
        } else
          print((A[i][p] - A[1][p]) / B[1][p]);
      else {
        rec[get(make_pair(B[1][p], B[i][p]))] =
            get(make_pair(A[1][p] - A[i][p], B[i][p]));
        if (A[i][p] - A[1][p] > 0LL) {
          if (B[1][p] == 0LL) GG();
          limit = max(limit, (A[i][p] - A[1][p]) / B[1][p]);
          while (B[1][p] * limit < A[i][p] - A[1][p]) ++limit;
        }
        insert(B[1][p], A[i][p] - A[1][p], B[i][p]);
      }
    }
    if (rec.size() >= 2) {
      pair<long long int, long long int> k1 = rec.begin()->first,
                                         b1 = rec.begin()->second;
      pair<long long int, long long int> k2 = rec.rbegin()->first,
                                         b2 = rec.rbegin()->second;
      pair<long long int, long long int> x = (b2 - b1) / (k1 - k2);
      if (x.second != 1LL) GG();
      print(x.first);
    }
  }
  long long int k = (limit - A1) / MOD1;
  k = max(k, 0LL);
  while (k * MOD1 + A1 < limit) ++k;
  print(k * MOD1 + A1);
  return 0;
}
