#include <bits/stdc++.h>
using namespace std;
string s, t;
long long p[1000010];
long long pw[1000010];
const int base = 129;
const int mod = 1000000000 + 7;
int n, m;
int Z, O;
int first;
inline long long value(int i, int j) {
  return (p[j] + mod - ((p[i - 1] * pw[j - i + 1]) % mod)) % mod;
}
bool good(int p) {
  if (n <= 1LL * p * Z) {
    return false;
  }
  int z = p;
  int o = (n - z * Z) / O;
  if (o <= 0 || z <= 0) return false;
  if ((o * O) + (z * Z) != n) return false;
  long long ho;
  long long hz;
  hz = value(1, z);
  ho = value(z * first + 1, z * first + o);
  if (hz == ho) return false;
  long long ans = 0;
  for (int i = 1; i <= m; i++) {
    if (t[i] == '1') {
      ans *= pw[o];
      ans += ho;
      ans %= mod;
    } else {
      ans *= pw[z];
      ans += hz;
      ans %= mod;
    }
  }
  return ans == value(1, n);
}
int main(int argc, char const *argv[]) {
  pw[0] = 1;
  for (int i = 1; i <= 1000000; i++) {
    pw[i] = pw[i - 1] * base;
    pw[i] %= mod;
  }
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> t >> s;
  n = s.size();
  m = t.size();
  s = "&" + s;
  t = "&" + t;
  for (int i = 1; i <= n; i++) {
    p[i] = p[i - 1] * base + s[i];
    p[i] %= mod;
  }
  if (t[1] == '1') {
    for (int i = 1; i <= m; i++) {
      if (t[i] == '1')
        t[i] = '0';
      else
        t[i] = '1';
    }
  }
  for (int i = 1; i <= m; i++) {
    if (t[i] == '0')
      ++Z;
    else
      ++O;
  }
  first = 1;
  for (int i = 2; i <= m; i++) {
    if (t[1] == t[i]) {
      ++first;
    } else {
      break;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += good(i);
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
string s, t;
long long p[1000010];
long long pw[1000010];
const int base = 129;
const int mod = 1000000000 + 7;
int n, m;
int Z, O;
int first;
inline long long value(int i, int j) {
  return (p[j] + mod - ((p[i - 1] * pw[j - i + 1]) % mod)) % mod;
}
bool good(int p) {
  if (n <= 1LL * p * Z) {
    return false;
  }
  int z = p;
  int o = (n - z * Z) / O;
  if (o <= 0 || z <= 0) return false;
  if ((o * O) + (z * Z) != n) return false;
  long long ho;
  long long hz;
  hz = value(1, z);
  ho = value(z * first + 1, z * first + o);
  if (hz == ho) return false;
  long long ans = 0;
  for (int i = 1; i <= m; i++) {
    if (t[i] == '1') {
      ans *= pw[o];
      ans += ho;
      ans %= mod;
    } else {
      ans *= pw[z];
      ans += hz;
      ans %= mod;
    }
  }
  return ans == value(1, n);
}
int main(int argc, char const *argv[]) {
  pw[0] = 1;
  for (int i = 1; i <= 1000000; i++) {
    pw[i] = pw[i - 1] * base;
    pw[i] %= mod;
  }
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> t >> s;
  n = s.size();
  m = t.size();
  s = "&" + s;
  t = "&" + t;
  for (int i = 1; i <= n; i++) {
    p[i] = p[i - 1] * base + s[i];
    p[i] %= mod;
  }
  if (t[1] == '1') {
    for (int i = 1; i <= m; i++) {
      if (t[i] == '1')
        t[i] = '0';
      else
        t[i] = '1';
    }
  }
  for (int i = 1; i <= m; i++) {
    if (t[i] == '0')
      ++Z;
    else
      ++O;
  }
  first = 1;
  for (int i = 2; i <= m; i++) {
    if (t[1] == t[i]) {
      ++first;
    } else {
      break;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += good(i);
  }
  cout << ans << '\n';
  return 0;
}

```