#include <bits/stdc++.h>
using namespace std;
long long hs[1000010], aoi[1000010] = {1};
inline long long sana(long long l, long long r) {
  return (hs[r] - hs[l - 1] * aoi[r - l + 1] % 100000007 + 100000007) %
         100000007;
}
inline bool check(long long a, long long b, string &s) {
  long long x, y, i;
  x = y = -1;
  i = 1;
  for (char c : s) {
    if (c == '0') {
      if (x == -1)
        x = sana(i, i + a - 1);
      else if (sana(i, i + a - 1) != x)
        return 0;
      i += a;
    } else {
      if (y == -1)
        y = sana(i, i + b - 1);
      else if (sana(i, i + b - 1) != y)
        return 0;
      i += b;
    }
  }
  return !(x == y);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, i, ans;
  a = b = ans = 0;
  string s, t;
  cin >> s;
  cin >> t;
  for (char c : s) {
    if (c == '0')
      a++;
    else
      b++;
  }
  for (i = 1; i <= t.size(); i++) {
    hs[i] = (hs[i - 1] * 29 + t[i - 1] - 'a' + 1) % 100000007;
    aoi[i] = aoi[i - 1] * 29 % 100000007;
  }
  for (i = 1; a * i < t.size(); i++) {
    if ((t.size() - a * i) % b == 0) {
      if (check(i, (t.size() - a * i) / b, s)) {
        ans++;
      }
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
long long hs[1000010], aoi[1000010] = {1};
inline long long sana(long long l, long long r) {
  return (hs[r] - hs[l - 1] * aoi[r - l + 1] % 100000007 + 100000007) %
         100000007;
}
inline bool check(long long a, long long b, string &s) {
  long long x, y, i;
  x = y = -1;
  i = 1;
  for (char c : s) {
    if (c == '0') {
      if (x == -1)
        x = sana(i, i + a - 1);
      else if (sana(i, i + a - 1) != x)
        return 0;
      i += a;
    } else {
      if (y == -1)
        y = sana(i, i + b - 1);
      else if (sana(i, i + b - 1) != y)
        return 0;
      i += b;
    }
  }
  return !(x == y);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, i, ans;
  a = b = ans = 0;
  string s, t;
  cin >> s;
  cin >> t;
  for (char c : s) {
    if (c == '0')
      a++;
    else
      b++;
  }
  for (i = 1; i <= t.size(); i++) {
    hs[i] = (hs[i - 1] * 29 + t[i - 1] - 'a' + 1) % 100000007;
    aoi[i] = aoi[i - 1] * 29 % 100000007;
  }
  for (i = 1; a * i < t.size(); i++) {
    if ((t.size() - a * i) % b == 0) {
      if (check(i, (t.size() - a * i) / b, s)) {
        ans++;
      }
    }
  }
  cout << ans << '\n';
  return 0;
}

```