#include <bits/stdc++.h>
int up(int x, int y) { return x / y + (x % y != 0); }
int main() {
  std::ios_base::sync_with_stdio(false), std::cin.tie(0);
  int A, B, n;
  std::cin >> A >> B;
  n = A + B;
  int ans = 0;
  for (int l = 1, r, t; l <= n; l = r + 1) {
    t = n / l, r = n / t;
    int ll = std::max(up(A, t + 1) + up(B, t + 1), l);
    int rr = std::min(A / t + B / t, r);
    if (up(A, t + 1) > A / t) continue;
    if (up(B, t + 1) > B / t) continue;
    ans += std::max(rr - ll + 1, 0);
  }
  std::cout << ans << std::endl;
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

You are given a uppercase Latin letters 'A' and b letters 'B'.

The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.

For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.

Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.

Input

The first line contains two integers a and b (1 ≤ a, b ≤ 10^9) — the number of letters 'A' and 'B', respectively.

Output

Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.

Examples

Input


2 4


Output


4


Input


5 3


Output


5

Note

All the possible periods for the first example: 

  * 3 "BBABBA" 
  * 4 "BBAABB" 
  * 5 "BBBAAB" 
  * 6 "AABBBB" 



All the possible periods for the second example: 

  * 3 "BAABAABA" 
  * 5 "BAABABAA" 
  * 6 "BABAAABA" 
  * 7 "BAABAAAB" 
  * 8 "AAAAABBB" 



Note that these are not the only possible strings for the given periods.

### Response

```cpp
#include <bits/stdc++.h>
int up(int x, int y) { return x / y + (x % y != 0); }
int main() {
  std::ios_base::sync_with_stdio(false), std::cin.tie(0);
  int A, B, n;
  std::cin >> A >> B;
  n = A + B;
  int ans = 0;
  for (int l = 1, r, t; l <= n; l = r + 1) {
    t = n / l, r = n / t;
    int ll = std::max(up(A, t + 1) + up(B, t + 1), l);
    int rr = std::min(A / t + B / t, r);
    if (up(A, t + 1) > A / t) continue;
    if (up(B, t + 1) > B / t) continue;
    ans += std::max(rr - ll + 1, 0);
  }
  std::cout << ans << std::endl;
  return 0;
}

```