#include <bits/stdc++.h>
using namespace std;
vector<long long> MOD = {1000000007, 1000000009, 998244353};
vector<long long> PRIMES = {179, 231, 119};
long long h[1000010][3];
long long pw[1000010][3];
long long K = 1000002;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s, t;
  cin >> s >> t;
  long long zr = 0, on = 0;
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] == '0')
      zr++;
    else
      on++;
  }
  for (long long i = 0; i < 3; i++) pw[0][i] = 1;
  for (long long i = 1; i < 1000010; i++) {
    for (long long j = 0; j < 3; j++)
      pw[i][j] = (pw[i - 1][j] * PRIMES[j]) % MOD[j];
  }
  if (on > zr) {
    swap(on, zr);
    for (long long i = 0; i < s.size(); i++) {
      if (s[i] == '0')
        s[i] = '1';
      else
        s[i] = '0';
    }
  }
  for (long long i = 0; i < 3; i++) h[0][i] = 0;
  for (long long i = 0; i < t.size(); i++) {
    for (long long j = 0; j < 3; j++) {
      h[i + 1][j] = (h[i][j] + pw[i][j] * (t[i] - 'a' + 1)) % MOD[j];
    }
  }
  long long z = 1;
  long long ans = 0;
  long long hz[3], ho[3];
  while (true) {
    if (z * zr >= t.size()) break;
    long long ost = t.size() - z * zr;
    if (ost % on != 0) {
      z++;
      continue;
    }
    long long o = ost / on;
    long long u = 0;
    for (long long i = 0; i < 3; i++) hz[i] = -1;
    for (long long i = 0; i < 3; i++) ho[i] = -1;
    bool ab = true;
    for (long long i = 0; i < s.size(); i++) {
      long long L, R;
      if (s[i] == '0') {
        L = u, R = u + z - 1;
        for (long long j = 0; j < 3; j++) {
          long long H = (h[R + 1][j] - h[L][j]) % MOD[j];
          H *= pw[K - R][j];
          H %= MOD[j];
          if (H < 0) H += MOD[j];
          if (hz[j] != -1 && hz[j] != H) ab = false;
          hz[j] = H;
        }
        u += z;
      } else {
        L = u, R = u + o - 1;
        for (long long j = 0; j < 3; j++) {
          long long H = (h[R + 1][j] - h[L][j]) % MOD[j];
          H *= pw[K - R][j];
          H %= MOD[j];
          if (H < 0) H += MOD[j];
          if (ho[j] != -1 && ho[j] != H) ab = false;
          ho[j] = H;
        }
        u += o;
      }
    }
    if (ho[0] == hz[0] && ho[1] == hz[1] && ho[2] == hz[2]) ab = false;
    if (ab) {
      ans++;
    }
    z++;
  }
  cout << ans;
}

### 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;
vector<long long> MOD = {1000000007, 1000000009, 998244353};
vector<long long> PRIMES = {179, 231, 119};
long long h[1000010][3];
long long pw[1000010][3];
long long K = 1000002;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s, t;
  cin >> s >> t;
  long long zr = 0, on = 0;
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] == '0')
      zr++;
    else
      on++;
  }
  for (long long i = 0; i < 3; i++) pw[0][i] = 1;
  for (long long i = 1; i < 1000010; i++) {
    for (long long j = 0; j < 3; j++)
      pw[i][j] = (pw[i - 1][j] * PRIMES[j]) % MOD[j];
  }
  if (on > zr) {
    swap(on, zr);
    for (long long i = 0; i < s.size(); i++) {
      if (s[i] == '0')
        s[i] = '1';
      else
        s[i] = '0';
    }
  }
  for (long long i = 0; i < 3; i++) h[0][i] = 0;
  for (long long i = 0; i < t.size(); i++) {
    for (long long j = 0; j < 3; j++) {
      h[i + 1][j] = (h[i][j] + pw[i][j] * (t[i] - 'a' + 1)) % MOD[j];
    }
  }
  long long z = 1;
  long long ans = 0;
  long long hz[3], ho[3];
  while (true) {
    if (z * zr >= t.size()) break;
    long long ost = t.size() - z * zr;
    if (ost % on != 0) {
      z++;
      continue;
    }
    long long o = ost / on;
    long long u = 0;
    for (long long i = 0; i < 3; i++) hz[i] = -1;
    for (long long i = 0; i < 3; i++) ho[i] = -1;
    bool ab = true;
    for (long long i = 0; i < s.size(); i++) {
      long long L, R;
      if (s[i] == '0') {
        L = u, R = u + z - 1;
        for (long long j = 0; j < 3; j++) {
          long long H = (h[R + 1][j] - h[L][j]) % MOD[j];
          H *= pw[K - R][j];
          H %= MOD[j];
          if (H < 0) H += MOD[j];
          if (hz[j] != -1 && hz[j] != H) ab = false;
          hz[j] = H;
        }
        u += z;
      } else {
        L = u, R = u + o - 1;
        for (long long j = 0; j < 3; j++) {
          long long H = (h[R + 1][j] - h[L][j]) % MOD[j];
          H *= pw[K - R][j];
          H %= MOD[j];
          if (H < 0) H += MOD[j];
          if (ho[j] != -1 && ho[j] != H) ab = false;
          ho[j] = H;
        }
        u += o;
      }
    }
    if (ho[0] == hz[0] && ho[1] == hz[1] && ho[2] == hz[2]) ab = false;
    if (ab) {
      ans++;
    }
    z++;
  }
  cout << ans;
}

```