#include <bits/stdc++.h>
const int N = 1e6 + 10;
const int INF = 1e9 + 9;
const unsigned long long MOD = 1e9 + 9;
const unsigned long long p = 29;
unsigned long long ppow[N];
unsigned long long h[N];
unsigned long long hash(int l, int r) {
  return (MOD + h[r] - h[l - 1] * ppow[r - l + 1] % MOD) % MOD;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  ppow[0] = 1;
  for (int i = 1; i < N; i++) {
    ppow[i] = ppow[i - 1] * p % MOD;
  }
  std::string s, t;
  std::cin >> s >> t;
  int cnt0 = 0, cnt1 = 0;
  for (char i : s) {
    if (i == '0')
      cnt0++;
    else
      cnt1++;
  }
  h[0] = 0;
  for (int i = 1; i <= t.length(); i++) {
    h[i] = (h[i - 1] * p % MOD + t[i - 1] - 'a' + 1) % MOD;
  }
  unsigned long long h0, h1;
  int ans = 0;
  for (int len0 = 1; len0 <= t.length(); len0++) {
    long long sum1 = (long long)t.length() - 1ll * len0 * cnt0;
    if (sum1 % cnt1 || sum1 <= 0) continue;
    int len1 = sum1 / cnt1;
    int pos = 1, i = 0;
    h0 = h1 = MOD;
    bool good = true;
    while (i < s.length() && good) {
      if (s[i] == '0') {
        if (h0 == MOD) {
          h0 = hash(pos, pos + len0 - 1);
        } else {
          if (h0 != hash(pos, pos + len0 - 1)) {
            good = false;
          }
          i++;
          pos += len0;
        }
      } else {
        if (h1 == MOD) {
          h1 = hash(pos, pos + len1 - 1);
        } else {
          if (h1 != hash(pos, pos + len1 - 1)) {
            good = false;
          }
          i++;
          pos += len1;
        }
      }
    }
    if (good && h0 != h1) {
      ans++;
    }
  }
  printf("%d", ans);
}

### Prompt

Your task is to create 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>
const int N = 1e6 + 10;
const int INF = 1e9 + 9;
const unsigned long long MOD = 1e9 + 9;
const unsigned long long p = 29;
unsigned long long ppow[N];
unsigned long long h[N];
unsigned long long hash(int l, int r) {
  return (MOD + h[r] - h[l - 1] * ppow[r - l + 1] % MOD) % MOD;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  ppow[0] = 1;
  for (int i = 1; i < N; i++) {
    ppow[i] = ppow[i - 1] * p % MOD;
  }
  std::string s, t;
  std::cin >> s >> t;
  int cnt0 = 0, cnt1 = 0;
  for (char i : s) {
    if (i == '0')
      cnt0++;
    else
      cnt1++;
  }
  h[0] = 0;
  for (int i = 1; i <= t.length(); i++) {
    h[i] = (h[i - 1] * p % MOD + t[i - 1] - 'a' + 1) % MOD;
  }
  unsigned long long h0, h1;
  int ans = 0;
  for (int len0 = 1; len0 <= t.length(); len0++) {
    long long sum1 = (long long)t.length() - 1ll * len0 * cnt0;
    if (sum1 % cnt1 || sum1 <= 0) continue;
    int len1 = sum1 / cnt1;
    int pos = 1, i = 0;
    h0 = h1 = MOD;
    bool good = true;
    while (i < s.length() && good) {
      if (s[i] == '0') {
        if (h0 == MOD) {
          h0 = hash(pos, pos + len0 - 1);
        } else {
          if (h0 != hash(pos, pos + len0 - 1)) {
            good = false;
          }
          i++;
          pos += len0;
        }
      } else {
        if (h1 == MOD) {
          h1 = hash(pos, pos + len1 - 1);
        } else {
          if (h1 != hash(pos, pos + len1 - 1)) {
            good = false;
          }
          i++;
          pos += len1;
        }
      }
    }
    if (good && h0 != h1) {
      ans++;
    }
  }
  printf("%d", ans);
}

```