#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 998244353;
const long long maxn = 2e5 + 107;
const double pi = acos(-1.0);
const double eps = 0.0000000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline void pr2(T x, int k = 64) {
  long long i;
  for (i = 0; i < k; i++) fprintf(stderr, "%d", (x >> i) & 1);
  putchar(' ');
}
template <typename T>
inline void add_(T &A, int B, long long MOD = M) {
  A += B;
  (A >= MOD) && (A -= MOD);
}
template <typename T>
inline void mul_(T &A, long long B, long long MOD = M) {
  A = (A * B) % MOD;
}
template <typename T>
inline void mod_(T &A, long long MOD = M) {
  A %= MOD;
  A += MOD;
  A %= MOD;
}
template <typename T>
inline void max_(T &A, T B) {
  (A < B) && (A = B);
}
template <typename T>
inline void min_(T &A, T B) {
  (A > B) && (A = B);
}
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
inline long long powMM(long long a, long long b, long long mod = M) {
  long long ret = 1;
  for (; b; b >>= 1ll, a = a * a % mod)
    if (b & 1) ret = ret * a % mod;
  return ret;
}
int startTime;
void startTimer() { startTime = clock(); }
void printTimer() {
  fprintf(stderr, "/--- Time: %ld milliseconds ---/\n", clock() - startTime);
}
int main() {
  long long a, b;
  long long ans = 0;
  scanf("%lld%lld", &a, &b);
  for (long long l = 1, r; l <= a + b; l = r + 1) {
    long long g = (a + b) / l;
    r = (a + b) / g;
    long long al = (a + g) / (g + 1), ar = a / g;
    long long bl = (b + g) / (g + 1), br = b / g;
    if (al > ar || bl > br) continue;
    long long right = min(r, ar + br);
    long long left = max(l, al + bl);
    if (right >= left) ans += right - left + 1;
  }
  printf("%lld\n", ans);
}

### Prompt

Your task is to create a CPP solution 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>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 998244353;
const long long maxn = 2e5 + 107;
const double pi = acos(-1.0);
const double eps = 0.0000000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline void pr2(T x, int k = 64) {
  long long i;
  for (i = 0; i < k; i++) fprintf(stderr, "%d", (x >> i) & 1);
  putchar(' ');
}
template <typename T>
inline void add_(T &A, int B, long long MOD = M) {
  A += B;
  (A >= MOD) && (A -= MOD);
}
template <typename T>
inline void mul_(T &A, long long B, long long MOD = M) {
  A = (A * B) % MOD;
}
template <typename T>
inline void mod_(T &A, long long MOD = M) {
  A %= MOD;
  A += MOD;
  A %= MOD;
}
template <typename T>
inline void max_(T &A, T B) {
  (A < B) && (A = B);
}
template <typename T>
inline void min_(T &A, T B) {
  (A > B) && (A = B);
}
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
inline long long powMM(long long a, long long b, long long mod = M) {
  long long ret = 1;
  for (; b; b >>= 1ll, a = a * a % mod)
    if (b & 1) ret = ret * a % mod;
  return ret;
}
int startTime;
void startTimer() { startTime = clock(); }
void printTimer() {
  fprintf(stderr, "/--- Time: %ld milliseconds ---/\n", clock() - startTime);
}
int main() {
  long long a, b;
  long long ans = 0;
  scanf("%lld%lld", &a, &b);
  for (long long l = 1, r; l <= a + b; l = r + 1) {
    long long g = (a + b) / l;
    r = (a + b) / g;
    long long al = (a + g) / (g + 1), ar = a / g;
    long long bl = (b + g) / (g + 1), br = b / g;
    if (al > ar || bl > br) continue;
    long long right = min(r, ar + br);
    long long left = max(l, al + bl);
    if (right >= left) ans += right - left + 1;
  }
  printf("%lld\n", ans);
}

```