#include <bits/stdc++.h>
using namespace std;
int a, b, c, d, e;
char s1[100007], s2[1000007];
long long h[1000007] = {0}, p[1000007], co = 1e9 + 7;
int main() {
  scanf("%s", s1);
  scanf("%s", s2);
  int l1 = strlen(s1), l2 = strlen(s2);
  p[0] = 1;
  for (int i = 1; i <= 1000005; i++) p[i] = p[i - 1] * 26 % co;
  for (int i = 1; i <= l2; i++)
    h[i] = h[i - 1] + p[i - 1] * (s2[i - 1] - 'a') % co;
  int n0 = 0, n1 = 0, ans = 0;
  for (int i = 1; i <= l1; i++)
    if (s1[i - 1] == s1[0])
      n0++;
    else
      n1++;
  for (int i = 1; i <= (l2 - 1) / n0; i++)
    if ((l2 - i * n0) % n1 == 0) {
      b = 0;
      c = (l2 - i * n0) / n1;
      int f = 1;
      long long ha = -1, hb = -1;
      int p2 = 0;
      for (int j = 0; j <= l1 - 1; j++)
        if (s1[j] == s1[0]) {
          if (ha == -1)
            ha = ((h[b + i] - h[b]) % co + co) % co;
          else if (ha * p[b] % co != ((h[b + i] - h[b]) % co + co) % co) {
            f = 0;
            break;
          }
          b += i;
        } else {
          if (hb == -1)
            hb = ((h[b + c] - h[b]) % co + co) % co, p2 = b;
          else if (hb * p[b - p2] % co != ((h[b + c] - h[b]) % co + co) % co) {
            f = 0;
            break;
          }
          b += c;
        }
      if (f && i == c && ha * p[p2] % co == hb) ans--;
      if (f) ans++;
    }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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;
int a, b, c, d, e;
char s1[100007], s2[1000007];
long long h[1000007] = {0}, p[1000007], co = 1e9 + 7;
int main() {
  scanf("%s", s1);
  scanf("%s", s2);
  int l1 = strlen(s1), l2 = strlen(s2);
  p[0] = 1;
  for (int i = 1; i <= 1000005; i++) p[i] = p[i - 1] * 26 % co;
  for (int i = 1; i <= l2; i++)
    h[i] = h[i - 1] + p[i - 1] * (s2[i - 1] - 'a') % co;
  int n0 = 0, n1 = 0, ans = 0;
  for (int i = 1; i <= l1; i++)
    if (s1[i - 1] == s1[0])
      n0++;
    else
      n1++;
  for (int i = 1; i <= (l2 - 1) / n0; i++)
    if ((l2 - i * n0) % n1 == 0) {
      b = 0;
      c = (l2 - i * n0) / n1;
      int f = 1;
      long long ha = -1, hb = -1;
      int p2 = 0;
      for (int j = 0; j <= l1 - 1; j++)
        if (s1[j] == s1[0]) {
          if (ha == -1)
            ha = ((h[b + i] - h[b]) % co + co) % co;
          else if (ha * p[b] % co != ((h[b + i] - h[b]) % co + co) % co) {
            f = 0;
            break;
          }
          b += i;
        } else {
          if (hb == -1)
            hb = ((h[b + c] - h[b]) % co + co) % co, p2 = b;
          else if (hb * p[b - p2] % co != ((h[b + c] - h[b]) % co + co) % co) {
            f = 0;
            break;
          }
          b += c;
        }
      if (f && i == c && ha * p[p2] % co == hb) ans--;
      if (f) ans++;
    }
  printf("%d\n", ans);
  return 0;
}

```