#include <bits/stdc++.h>
using namespace std;
const int max_n = 1000005;
const long long mod = 1e9 + 7;
const int seed = 233;
char s[max_n], t[max_n];
long long hs[max_n], p[max_n];
void shash(char *s) {
  p[0] = 1;
  int len = strlen(s + 1);
  for (int i = 1; i <= len; i++) {
    p[i] = p[i - 1] * seed % mod;
    hs[i] = (hs[i - 1] * seed + s[i]) % mod;
  }
}
bool check(int x, int y, char c) {
  long long h1 = -1, h2 = -1;
  long long tl = 0, th;
  int l, r;
  for (int i = 1; s[i]; i++) {
    if (s[i] == c) {
      l = tl + 1, r = tl + 1 + x - 1;
      th = hs[r] + mod - hs[l - 1] * p[x] % mod, th %= mod;
      if (h1 == -1)
        h1 = th;
      else if (h1 != th)
        return false;
      tl += x;
    } else {
      l = tl + 1, r = tl + 1 + y - 1;
      th = hs[r] + mod - hs[l - 1] * p[y] % mod, th %= mod;
      if (h2 == -1)
        h2 = th;
      else if (h2 != th)
        return false;
      tl += y;
    }
  }
  if (h1 == h2) return false;
  return true;
}
int main() {
  scanf("%s%s", s + 1, t + 1);
  int len1 = strlen(s + 1), len2 = strlen(t + 1);
  int cnt = 0;
  char c = '0';
  for (int i = 1; i <= len1; i++)
    if (s[i] == '0') cnt++;
  if (cnt * 2 < len1) cnt = len1 - cnt, c = '1';
  shash(t);
  int ans = 0;
  for (int i = 1; i <= len2; i++) {
    if (1ll * i * cnt >= len2) break;
    if ((len2 - 1ll * i * cnt) % (len1 - cnt)) continue;
    if (check(i, (len2 - 1ll * i * cnt) / (len1 - cnt), c)) ans++;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

One of Arkady's friends works at a huge radio telescope. A few decades ago the telescope has sent a signal s towards a faraway galaxy. Recently they've received a response t which they believe to be a response from aliens! The scientists now want to check if the signal t is similar to s.

The original signal s was a sequence of zeros and ones (everyone knows that binary code is the universe-wide language). The returned signal t, however, does not look as easy as s, but the scientists don't give up! They represented t as a sequence of English letters and say that t is similar to s if you can replace all zeros in s with some string r_0 and all ones in s with some other string r_1 and obtain t. The strings r_0 and r_1 must be different and non-empty.

Please help Arkady's friend and find the number of possible replacements for zeros and ones (the number of pairs of strings r_0 and r_1) that transform s to t.

Input

The first line contains a string s (2 ≤ |s| ≤ 10^5) consisting of zeros and ones — the original signal.

The second line contains a string t (1 ≤ |t| ≤ 10^6) consisting of lowercase English letters only — the received signal.

It is guaranteed, that the string s contains at least one '0' and at least one '1'.

Output

Print a single integer — the number of pairs of strings r_0 and r_1 that transform s to t.

In case there are no such pairs, print 0.

Examples

Input


01
aaaaaa


Output


4


Input


001
kokokokotlin


Output


2

Note

In the first example, the possible pairs (r_0, r_1) are as follows:

  * "a", "aaaaa" 
  * "aa", "aaaa" 
  * "aaaa", "aa" 
  * "aaaaa", "a" 



The pair "aaa", "aaa" is not allowed, since r_0 and r_1 must be different.

In the second example, the following pairs are possible: 

  * "ko", "kokotlin" 
  * "koko", "tlin" 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int max_n = 1000005;
const long long mod = 1e9 + 7;
const int seed = 233;
char s[max_n], t[max_n];
long long hs[max_n], p[max_n];
void shash(char *s) {
  p[0] = 1;
  int len = strlen(s + 1);
  for (int i = 1; i <= len; i++) {
    p[i] = p[i - 1] * seed % mod;
    hs[i] = (hs[i - 1] * seed + s[i]) % mod;
  }
}
bool check(int x, int y, char c) {
  long long h1 = -1, h2 = -1;
  long long tl = 0, th;
  int l, r;
  for (int i = 1; s[i]; i++) {
    if (s[i] == c) {
      l = tl + 1, r = tl + 1 + x - 1;
      th = hs[r] + mod - hs[l - 1] * p[x] % mod, th %= mod;
      if (h1 == -1)
        h1 = th;
      else if (h1 != th)
        return false;
      tl += x;
    } else {
      l = tl + 1, r = tl + 1 + y - 1;
      th = hs[r] + mod - hs[l - 1] * p[y] % mod, th %= mod;
      if (h2 == -1)
        h2 = th;
      else if (h2 != th)
        return false;
      tl += y;
    }
  }
  if (h1 == h2) return false;
  return true;
}
int main() {
  scanf("%s%s", s + 1, t + 1);
  int len1 = strlen(s + 1), len2 = strlen(t + 1);
  int cnt = 0;
  char c = '0';
  for (int i = 1; i <= len1; i++)
    if (s[i] == '0') cnt++;
  if (cnt * 2 < len1) cnt = len1 - cnt, c = '1';
  shash(t);
  int ans = 0;
  for (int i = 1; i <= len2; i++) {
    if (1ll * i * cnt >= len2) break;
    if ((len2 - 1ll * i * cnt) % (len1 - cnt)) continue;
    if (check(i, (len2 - 1ll * i * cnt) / (len1 - cnt), c)) ans++;
  }
  printf("%d\n", ans);
  return 0;
}

```