#include <bits/stdc++.h>
using namespace std;
char str1[1000005], str2[1000005];
int cnt[3];
unsigned long long bao[1000005], base[1000005];
long long base2[1000005], da[1000005];
int main() {
  int n, m, i, j, ans = 0;
  scanf("%s", str1 + 1);
  scanf("%s", str2 + 1);
  n = strlen(str1 + 1);
  m = strlen(str2 + 1);
  base[0] = 1;
  base2[0] = 1;
  for (i = 1; i <= m; i++) {
    base[i] = base[i - 1] * 233;
    bao[i] = bao[i - 1] * 233 + str2[i] - 'a' + 1;
    base2[i] = base2[i - 1] * 2333 % 1000000007;
    da[i] = (da[i - 1] * 2333 + str2[i] - 'a' + 1) % 1000000007;
  }
  for (i = 1; i <= n; i++) {
    cnt[str1[i] - '0']++;
  }
  int f = str1[1] - '0';
  for (i = 1; i <= m / cnt[f]; i++) {
    int w = m - cnt[f] * i;
    if (w % cnt[1 - f] != 0 || w == 0) continue;
    w /= cnt[1 - f];
    unsigned long long a1 = bao[i], a2 = 114514;
    long long b1 = da[i], b2 = 114514;
    int now = i;
    for (j = 2; j <= n; j++) {
      if (str1[j] == f + '0') {
        if (a1 != bao[now + i] - bao[now] * base[i]) break;
        if (b1 !=
            (da[now + i] - 1LL * da[now] * base2[i] % 1000000007 + 1000000007) %
                1000000007)
          break;
        now += i;
      } else {
        if (a2 == 114514 && b2 == 114514) {
          a2 = bao[now + w] - bao[now] * base[w];
          b2 = (da[now + w] - 1LL * da[now] * base2[w] % 1000000007 +
                1000000007) %
               1000000007;
        } else {
          if (a2 != bao[now + w] - bao[now] * base[w]) break;
          if (b2 != (da[now + w] - 1LL * da[now] * base2[w] % 1000000007 +
                     1000000007) %
                        1000000007)
            break;
        }
        now += w;
      }
    }
    if (j > n) {
      if (a1 != a2 && b1 != b2) ans++;
    }
  }
  cout << ans << endl;
}

### Prompt

Create a solution in Cpp for 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;
char str1[1000005], str2[1000005];
int cnt[3];
unsigned long long bao[1000005], base[1000005];
long long base2[1000005], da[1000005];
int main() {
  int n, m, i, j, ans = 0;
  scanf("%s", str1 + 1);
  scanf("%s", str2 + 1);
  n = strlen(str1 + 1);
  m = strlen(str2 + 1);
  base[0] = 1;
  base2[0] = 1;
  for (i = 1; i <= m; i++) {
    base[i] = base[i - 1] * 233;
    bao[i] = bao[i - 1] * 233 + str2[i] - 'a' + 1;
    base2[i] = base2[i - 1] * 2333 % 1000000007;
    da[i] = (da[i - 1] * 2333 + str2[i] - 'a' + 1) % 1000000007;
  }
  for (i = 1; i <= n; i++) {
    cnt[str1[i] - '0']++;
  }
  int f = str1[1] - '0';
  for (i = 1; i <= m / cnt[f]; i++) {
    int w = m - cnt[f] * i;
    if (w % cnt[1 - f] != 0 || w == 0) continue;
    w /= cnt[1 - f];
    unsigned long long a1 = bao[i], a2 = 114514;
    long long b1 = da[i], b2 = 114514;
    int now = i;
    for (j = 2; j <= n; j++) {
      if (str1[j] == f + '0') {
        if (a1 != bao[now + i] - bao[now] * base[i]) break;
        if (b1 !=
            (da[now + i] - 1LL * da[now] * base2[i] % 1000000007 + 1000000007) %
                1000000007)
          break;
        now += i;
      } else {
        if (a2 == 114514 && b2 == 114514) {
          a2 = bao[now + w] - bao[now] * base[w];
          b2 = (da[now + w] - 1LL * da[now] * base2[w] % 1000000007 +
                1000000007) %
               1000000007;
        } else {
          if (a2 != bao[now + w] - bao[now] * base[w]) break;
          if (b2 != (da[now + w] - 1LL * da[now] * base2[w] % 1000000007 +
                     1000000007) %
                        1000000007)
            break;
        }
        now += w;
      }
    }
    if (j > n) {
      if (a1 != a2 && b1 != b2) ans++;
    }
  }
  cout << ans << endl;
}

```