#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long xx = 0, flagg = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') {
    flagg = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    xx = xx * 10 + ch - '0';
    ch = getchar();
  }
  return xx * flagg;
}
void pus(long long xx, long long flagg) {
  if (xx < 0) {
    putchar('-');
    xx = -xx;
  }
  if (xx >= 10) pus(xx / 10, 0);
  putchar(xx % 10 + '0');
  if (flagg == 1) putchar(' ');
  if (flagg == 2) putchar('\n');
  return;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long g = exgcd(b, a % b, x, y), t = x;
  x = y;
  y = t - a / b * y;
  return g;
}
long long ksm(long long u, long long v) {
  long long o = 1;
  u %= 1000000007;
  while (v) {
    if (v & 1) o = o * u % 1000000007;
    u = u * u % 1000000007;
    v >>= 1;
  }
  return o;
}
long long n, p[2005], pn, a[105][2005], b[105][2005];
struct equation {
  long long A, B, C;
  equation() : A(0), B(0), C(0) {}
  equation(long long newA, long long newB, long long newC)
      : A(newA), B(newB), C(newC) {}
};
long long empty(equation &t) {
  if (t.A == 0 && t.B == 0) return t.C;
  long long g = gcd(abs(t.A), abs(t.B));
  if (t.C % g) return 1;
  t.A /= g;
  t.B /= g;
  t.C /= g;
  return 0;
}
long long merge(equation p, equation q, long long &x, long long &y) {
  long long D = p.A * q.B - p.B * q.A;
  x = p.C * q.B - q.C * p.B, y = p.A * q.C - p.C * q.A;
  if (D) {
    if (x % D || y % D) {
      x = -1;
      y = -1;
      return -1;
    }
    x /= D;
    y /= D;
    return 1;
  }
  if (x || y) {
    x = -1;
    y = -1;
    return -1;
  }
  return 0;
}
long long check(long long x) {
  if (x < 0) return -1;
  long long ans = 1, i, u, z;
  for (i = 2; i <= n; i++) {
    vector<long long> s;
    s.clear();
    for (u = 1; u <= pn; u++) {
      z = a[1][u] * x + b[1][u] - b[i][u];
      if (a[i][u]) {
        if (z % a[i][u]) return -1;
        s.push_back(z / a[i][u]);
      } else if (z)
        return -1;
    }
    sort(s.begin(), s.end());
    s.erase(unique(s.begin(), s.end()), s.end());
    if (s.size() > 1 || (s.size() == 1 && s.front() < 0)) return -1;
  }
  for (i = 1; i <= pn; i++)
    ans = ans * ksm(p[i], a[1][i] * x + b[1][i]) % 1000000007;
  return ans;
}
long long i, j, x, y, u[105], v[105], c[105], d[105], m, minx, A, B, C, g, t,
    flag;
int main() {
  n = read();
  for (i = 1; i <= n; i++) {
    u[i] = read();
    v[i] = read();
    for (x = u[i], y = 2; y <= x / y; y++)
      if (x % y == 0)
        for (p[++pn] = y; x % y == 0; x /= y)
          ;
    if (x != 1) p[++pn] = x;
    for (x = v[i], y = 2; y <= x / y; y++)
      if (x % y == 0)
        for (p[++pn] = y; x % y == 0; x /= y)
          ;
    if (x != 1) p[++pn] = x;
  }
  if (n == 1) {
    pus(u[1], 2);
    return 0;
  }
  sort(p + 1, p + pn + 1);
  pn = unique(p + 1, p + pn + 1) - p - 1;
  memset(a, 0, sizeof(a));
  memset(b, 0, sizeof(b));
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= pn; j++)
      while (u[i] % p[j] == 0) {
        b[i][j]++;
        u[i] /= p[j];
      }
    for (j = 1; j <= pn; j++)
      while (v[i] % p[j] == 0) {
        a[i][j]++;
        v[i] /= p[j];
      }
  }
  x = 0;
  y = 0;
  long long u, v;
  for (i = 2; flag == 0 && i <= n; i++)
    for (u = 1; flag == 0 && u <= pn; u++) {
      equation p(a[1][u], -a[i][u], b[i][u] - b[1][u]);
      if (empty(p)) {
        pus(-1, 2);
        return 0;
      }
      if (p.B == 0 && p.A) {
        x = p.C / p.A;
        flag = 1;
        break;
      }
      for (v = u + 1; v <= pn; v++) {
        equation q(a[1][v], -a[i][v], b[i][v] - b[1][v]);
        t = merge(p, q, x, y);
        if (t == 1) {
          flag = 1;
          break;
        }
        if (t == -1) {
          pus(-1, 2);
          return 0;
        }
      }
    }
  if (flag) {
    pus(check(x), 2);
    return 0;
  }
  for (i = 2; i <= n; i++)
    for (u = 1; u <= pn; u++)
      if (a[1][u]) {
        A = a[1][u];
        B = a[i][u];
        C = b[i][u] - b[1][u];
        g = exgcd(A, B, x, y);
        m++;
        minx = max(minx, (long long)ceil(1.0 * C / A));
        c[m] = (C / g) * x % B;
        d[m] = B / g;
        break;
      }
  for (i = 2; i <= m; i++) {
    A = d[1];
    B = d[i];
    C = c[i] - c[1];
    g = exgcd(A, B, x, y);
    if (C % g) {
      pus(-1, 2);
      return 0;
    }
    x = x * (C / g) % B;
    c[1] = c[1] + x * A;
    d[1] = A / g * B;
    c[1] %= d[1];
    if (c[1] < 0) c[1] += d[1];
  }
  c[1] += ceil(1.0 * (minx - c[1]) / d[1]);
  pus(check(c[1]), 2);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long read() {
  long long xx = 0, flagg = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') {
    flagg = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    xx = xx * 10 + ch - '0';
    ch = getchar();
  }
  return xx * flagg;
}
void pus(long long xx, long long flagg) {
  if (xx < 0) {
    putchar('-');
    xx = -xx;
  }
  if (xx >= 10) pus(xx / 10, 0);
  putchar(xx % 10 + '0');
  if (flagg == 1) putchar(' ');
  if (flagg == 2) putchar('\n');
  return;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long g = exgcd(b, a % b, x, y), t = x;
  x = y;
  y = t - a / b * y;
  return g;
}
long long ksm(long long u, long long v) {
  long long o = 1;
  u %= 1000000007;
  while (v) {
    if (v & 1) o = o * u % 1000000007;
    u = u * u % 1000000007;
    v >>= 1;
  }
  return o;
}
long long n, p[2005], pn, a[105][2005], b[105][2005];
struct equation {
  long long A, B, C;
  equation() : A(0), B(0), C(0) {}
  equation(long long newA, long long newB, long long newC)
      : A(newA), B(newB), C(newC) {}
};
long long empty(equation &t) {
  if (t.A == 0 && t.B == 0) return t.C;
  long long g = gcd(abs(t.A), abs(t.B));
  if (t.C % g) return 1;
  t.A /= g;
  t.B /= g;
  t.C /= g;
  return 0;
}
long long merge(equation p, equation q, long long &x, long long &y) {
  long long D = p.A * q.B - p.B * q.A;
  x = p.C * q.B - q.C * p.B, y = p.A * q.C - p.C * q.A;
  if (D) {
    if (x % D || y % D) {
      x = -1;
      y = -1;
      return -1;
    }
    x /= D;
    y /= D;
    return 1;
  }
  if (x || y) {
    x = -1;
    y = -1;
    return -1;
  }
  return 0;
}
long long check(long long x) {
  if (x < 0) return -1;
  long long ans = 1, i, u, z;
  for (i = 2; i <= n; i++) {
    vector<long long> s;
    s.clear();
    for (u = 1; u <= pn; u++) {
      z = a[1][u] * x + b[1][u] - b[i][u];
      if (a[i][u]) {
        if (z % a[i][u]) return -1;
        s.push_back(z / a[i][u]);
      } else if (z)
        return -1;
    }
    sort(s.begin(), s.end());
    s.erase(unique(s.begin(), s.end()), s.end());
    if (s.size() > 1 || (s.size() == 1 && s.front() < 0)) return -1;
  }
  for (i = 1; i <= pn; i++)
    ans = ans * ksm(p[i], a[1][i] * x + b[1][i]) % 1000000007;
  return ans;
}
long long i, j, x, y, u[105], v[105], c[105], d[105], m, minx, A, B, C, g, t,
    flag;
int main() {
  n = read();
  for (i = 1; i <= n; i++) {
    u[i] = read();
    v[i] = read();
    for (x = u[i], y = 2; y <= x / y; y++)
      if (x % y == 0)
        for (p[++pn] = y; x % y == 0; x /= y)
          ;
    if (x != 1) p[++pn] = x;
    for (x = v[i], y = 2; y <= x / y; y++)
      if (x % y == 0)
        for (p[++pn] = y; x % y == 0; x /= y)
          ;
    if (x != 1) p[++pn] = x;
  }
  if (n == 1) {
    pus(u[1], 2);
    return 0;
  }
  sort(p + 1, p + pn + 1);
  pn = unique(p + 1, p + pn + 1) - p - 1;
  memset(a, 0, sizeof(a));
  memset(b, 0, sizeof(b));
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= pn; j++)
      while (u[i] % p[j] == 0) {
        b[i][j]++;
        u[i] /= p[j];
      }
    for (j = 1; j <= pn; j++)
      while (v[i] % p[j] == 0) {
        a[i][j]++;
        v[i] /= p[j];
      }
  }
  x = 0;
  y = 0;
  long long u, v;
  for (i = 2; flag == 0 && i <= n; i++)
    for (u = 1; flag == 0 && u <= pn; u++) {
      equation p(a[1][u], -a[i][u], b[i][u] - b[1][u]);
      if (empty(p)) {
        pus(-1, 2);
        return 0;
      }
      if (p.B == 0 && p.A) {
        x = p.C / p.A;
        flag = 1;
        break;
      }
      for (v = u + 1; v <= pn; v++) {
        equation q(a[1][v], -a[i][v], b[i][v] - b[1][v]);
        t = merge(p, q, x, y);
        if (t == 1) {
          flag = 1;
          break;
        }
        if (t == -1) {
          pus(-1, 2);
          return 0;
        }
      }
    }
  if (flag) {
    pus(check(x), 2);
    return 0;
  }
  for (i = 2; i <= n; i++)
    for (u = 1; u <= pn; u++)
      if (a[1][u]) {
        A = a[1][u];
        B = a[i][u];
        C = b[i][u] - b[1][u];
        g = exgcd(A, B, x, y);
        m++;
        minx = max(minx, (long long)ceil(1.0 * C / A));
        c[m] = (C / g) * x % B;
        d[m] = B / g;
        break;
      }
  for (i = 2; i <= m; i++) {
    A = d[1];
    B = d[i];
    C = c[i] - c[1];
    g = exgcd(A, B, x, y);
    if (C % g) {
      pus(-1, 2);
      return 0;
    }
    x = x * (C / g) % B;
    c[1] = c[1] + x * A;
    d[1] = A / g * B;
    c[1] %= d[1];
    if (c[1] < 0) c[1] += d[1];
  }
  c[1] += ceil(1.0 * (minx - c[1]) / d[1]);
  pus(check(c[1]), 2);
  return 0;
}

```