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;
int n;
int a[110], b[110], c[110], d[110];
vector<int> stdprimes;
vector<int> sepra(int x) {
  vector<int> res = vector<int>();
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      res.push_back(i);
      while (x % i == 0) x /= i;
    }
  }
  if (x != 1) res.push_back(x);
  return res;
}
struct record {
  vector<int> stdpows;
  vector<int> extrapr;
  vector<int> extrapo;
  record() {
    stdpows.clear();
    extrapr.clear();
    extrapo.clear();
  }
  record(int x) {
    for (int i = 0; i < stdprimes.size(); i++) {
      stdpows.push_back(0);
      while (x % stdprimes[i] == 0) {
        x /= stdprimes[i];
        stdpows.back()++;
      }
    }
    for (int i = 2; i * i <= x; i++) {
      if (x % i == 0) {
        extrapr.push_back(i);
        extrapo.push_back(0);
        while (x % i == 0) {
          x /= i;
          extrapo.back()++;
        }
      }
    }
    if (x != 1) {
      extrapr.push_back(x);
      extrapo.push_back(1);
    }
  }
};
inline long long quickpow(long long base, long long ind) {
  long long ans = 1;
  while (ind > 0) {
    if (ind & 1) ans = ans * base % 1000000007;
    ind >>= 1;
    base = base * base % 1000000007;
  }
  return ans;
}
inline void mul(long long &a, long long b) { a = a * b % 1000000007; }
inline long long extgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  long long g = extgcd(b, a % b, y, x);
  y = y - a / b * x;
  return g;
}
void combine(long long &a1, long long &m1, int a2, int m2) {
  long long c = a2 - a1;
  long long a = m1, b = m2;
  long long x, y, g;
  g = extgcd(a, b, x, y);
  if (c % g != 0) {
    cout << -1;
    exit(0);
  }
  long long b1 = b / g;
  long long x1 = ((x + b1) * c / g);
  x1 = (x1 % b1 + b1) % b1;
  long long y1 = ((c - a * x1) / b);
  a1 = a1 + m1 * x1;
  m1 = m1 * m2 / g;
}
void trysingle(int id, int pos) {
  stdprimes.clear();
  record fst(a[id]);
  stdprimes = fst.extrapr;
  record ratio(b[id]);
  for (int i = 0; i < ratio.extrapr.size(); i++)
    stdprimes.push_back(ratio.extrapr[i]);
  fst = record(a[id]);
  ratio = record(b[id]);
  vector<int> cnt;
  for (int i = 0; i < fst.stdpows.size(); i++) {
    cnt.push_back(fst.stdpows[i]);
    cnt.back() += ratio.stdpows[i] * pos;
  }
  for (int i = 0; i < n; i++) {
    record curfst(a[i]);
    record currat(b[i]);
    if (!curfst.extrapr.empty()) {
      cout << -1;
      exit(0);
    }
    int curneed = -1;
    for (int j = 0; j < stdprimes.size(); j++) {
      if (curfst.stdpows[j] > cnt[j]) {
        cout << -1;
        exit(0);
      }
      int need = cnt[j] - curfst.stdpows[j];
      if (!currat.stdpows[j]) {
        if (need) {
          cout << -1;
          exit(0);
        }
        continue;
      } else if (need % currat.stdpows[j] != 0) {
        cout << -1;
        exit(0);
      }
      need /= currat.stdpows[j];
      if (curneed == -1)
        curneed = need;
      else if (curneed != need) {
        cout << -1;
        exit(0);
      }
    }
    if (curneed > 0 && !currat.extrapr.empty()) {
      cout << -1;
      exit(0);
    }
  }
  long long ans = a[id];
  mul(ans, quickpow(b[id], pos));
  cout << ans;
  exit(0);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i] >> b[i];
  for (int i = 0; i < n; i++)
    if (b[i] == 1) trysingle(i, 0);
  if (n == 1) {
    cout << a[0];
    return 0;
  }
  stdprimes = sepra(b[0]);
  record fst(a[0]);
  record ratio(b[0]);
  c[0] = fst.stdpows[0];
  d[0] = ratio.stdpows[0];
  for (int i = 1; i < n; i++) {
    record curfst(a[i]);
    record currat(b[i]);
    if (!currat.extrapr.empty()) {
      int needPow = 0, p = currat.extrapr[0], pw = currat.extrapo[0];
      for (int j = 0; j < fst.extrapr.size(); j++) {
        if (fst.extrapr[j] == p) {
          needPow += fst.extrapo[j];
          break;
        }
      }
      for (int j = 0; j < curfst.extrapr.size(); j++) {
        if (curfst.extrapr[j] == p) {
          needPow -= curfst.extrapo[j];
          break;
        }
      }
      if (needPow < 0 || needPow % pw != 0) {
        cout << -1;
        return 0;
      }
      trysingle(i, needPow / pw);
    }
    c[i] = curfst.stdpows[0];
    d[i] = currat.stdpows[0];
    int fstpos = 0, errpos = -1;
    while (fstpos < stdprimes.size() && currat.stdpows[fstpos] == 0) fstpos++;
    int pow0 = ratio.stdpows[fstpos];
    int powi = currat.stdpows[fstpos];
    for (int j = 0; j < stdprimes.size(); j++) {
      if (ratio.stdpows[j] * powi != currat.stdpows[j] * pow0) {
        errpos = j;
        break;
      }
    }
    if (~errpos) {
      int s1 = fst.stdpows[fstpos], s2 = fst.stdpows[errpos];
      int t1 = curfst.stdpows[fstpos], t2 = curfst.stdpows[errpos];
      int u1 = ratio.stdpows[fstpos], u2 = ratio.stdpows[errpos];
      int v1 = currat.stdpows[fstpos], v2 = currat.stdpows[errpos];
      int mulx = u1 * v2 - u2 * v1;
      int eqa = t1 * v2 - t2 * v1 - s1 * v2 + s2 * v1;
      if (mulx == 0 || (abs(eqa) % abs(mulx) != 0) || eqa / mulx < 0) {
        cout << -1;
        return 0;
      }
      trysingle(0, eqa / mulx);
    }
    if (fst.extrapr.size() != curfst.extrapr.size()) {
      cout << -1;
      return 0;
    }
    for (int j = 0; j < fst.extrapr.size(); j++)
      if (fst.extrapr[j] != curfst.extrapr[j] ||
          fst.extrapo[j] != curfst.extrapo[j]) {
        cout << -1;
        return 0;
      }
    int dif = curfst.stdpows[0] - fst.stdpows[0];
    for (int j = 1; j < stdprimes.size(); j++) {
      int curdif = curfst.stdpows[j] - fst.stdpows[j];
      if (dif * ratio.stdpows[j] != curdif * ratio.stdpows[0]) {
        cout << -1;
        return 0;
      }
    }
  }
  long long ansC = c[0], ansD = d[0];
  for (int i = 1; i < n; i++) {
    combine(ansC, ansD, c[i], d[i]);
  }
  for (int i = 0; i < n; i++) {
    if (ansC < c[i]) {
      (ansC += ((c[i] - ansC + ansD - 1) / ansD) * ansD);
    }
  }
  long long ans = a[0];
  ansC /= ratio.stdpows[0];
  mul(ans, quickpow(b[0], ansC));
  cout << ans;
  return 0;
}
