#include <bits/stdc++.h>
int main() {
  long long int p, q, count = 0, count1 = 0, n, j, req = 0, g;
  scanf("%lld%lld", &p, &q);
  scanf("%lld", &n);
  long long int a[n];
  for (j = 0; j < n; j++) {
    scanf("%lld", &a[j]);
  }
  g = n;
  if (p % q == 0 && a[1] == 1 && n == 2) {
    a[0] = a[0] + 1;
    g = 1;
  }
  if (p % q != 0 && a[1] == 1 && n == 2) {
    g = 2;
    a[0] = a[0];
    a[1] = a[1];
  }
  if (n != 2 && a[n - 1] == 1) {
    g = n - 1;
    a[n - 2] = a[n - 2] + 1;
  }
  if (p / q == a[0]) {
    long long int k, c[n], temp;
    for (k = 0;; k++) {
      c[k] = p / q;
      temp = q;
      q = (p - c[k] * q);
      p = temp;
      req++;
      if (q == 0) break;
    }
    for (k = 0; (k < req || k < g); k++) {
      if (c[k] == a[k]) count++;
    }
  }
  if (req > g) req = req;
  if (g > req) req = g;
  if (count == req && req != 0)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

A continued fraction of height n is a fraction of form <image>. You are given two rational numbers, one is represented as <image> and the other one is represented as a finite fraction of height n. Check if they are equal.

Input

The first line contains two space-separated integers p, q (1 ≤ q ≤ p ≤ 1018) — the numerator and the denominator of the first fraction.

The second line contains integer n (1 ≤ n ≤ 90) — the height of the second fraction. The third line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1018) — the continued fraction.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

Print "YES" if these fractions are equal and "NO" otherwise.

Examples

Input

9 4
2
2 4


Output

YES


Input

9 4
3
2 3 1


Output

YES


Input

9 4
3
1 2 4


Output

NO

Note

In the first sample <image>.

In the second sample <image>.

In the third sample <image>.

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  long long int p, q, count = 0, count1 = 0, n, j, req = 0, g;
  scanf("%lld%lld", &p, &q);
  scanf("%lld", &n);
  long long int a[n];
  for (j = 0; j < n; j++) {
    scanf("%lld", &a[j]);
  }
  g = n;
  if (p % q == 0 && a[1] == 1 && n == 2) {
    a[0] = a[0] + 1;
    g = 1;
  }
  if (p % q != 0 && a[1] == 1 && n == 2) {
    g = 2;
    a[0] = a[0];
    a[1] = a[1];
  }
  if (n != 2 && a[n - 1] == 1) {
    g = n - 1;
    a[n - 2] = a[n - 2] + 1;
  }
  if (p / q == a[0]) {
    long long int k, c[n], temp;
    for (k = 0;; k++) {
      c[k] = p / q;
      temp = q;
      q = (p - c[k] * q);
      p = temp;
      req++;
      if (q == 0) break;
    }
    for (k = 0; (k < req || k < g); k++) {
      if (c[k] == a[k]) count++;
    }
  }
  if (req > g) req = req;
  if (g > req) req = g;
  if (count == req && req != 0)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

```