#include <bits/stdc++.h>
using namespace std;
const int p = 233;
const int z = 998244353;
int fp(int a, int x) {
  if (x == 0) {
    return 1;
  }
  if (x % 2 == 1) {
    return 1ll * a * fp(a, x - 1) % z;
  }
  int t = fp(a, x / 2);
  return 1ll * t * t % z;
}
int inv(int x) { return fp(x, z - 2); }
int main() {
  ios::sync_with_stdio(false);
  string s, t;
  cin >> s;
  cin >> t;
  if (s[0] == '1') {
    for (int i = 0; i < s.length(); i++) {
      s[i] = '0' + '1' - s[i];
    }
  }
  int c0 = 0;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '0') {
      c0++;
    }
  }
  int c1 = s.length() - c0;
  int s1;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '1') {
      s1 = i;
      break;
    }
  }
  int f[t.length() + 1];
  f[0] = 0;
  int q = 1;
  for (int i = 0; i < t.length(); i++) {
    int x = t[i] - 'a';
    f[i + 1] = (f[i] + 1ll * x * q) % z;
    q = 1ll * q * p % z;
  }
  int invp[t.length()];
  q = inv(p);
  invp[0] = 1;
  for (int i = 1; i < t.length(); i++) {
    invp[i] = 1ll * invp[i - 1] * q % z;
  }
  int ans = 0;
  for (int i = 1; i < t.length(); i++) {
    if (c0 * i >= t.length()) {
      break;
    }
    if ((t.length() - c0 * i) % c1 != 0) {
      continue;
    }
    int l0 = i, l1 = (t.length() - c0 * i) / c1;
    int h0 = f[l0];
    int h1 = 1ll * (f[s1 * l0 + l1] + z - f[s1 * l0]) % z * invp[s1 * l0] % z;
    if (l0 == l1 && h0 == h1) {
      continue;
    }
    int k = 0;
    bool flag = true;
    for (int j = 0; j < s.length(); j++) {
      if (s[j] == '0') {
        int h = 1ll * (f[k + l0] + z - f[k]) % z * invp[k] % z;
        if (h != h0) {
          flag = false;
          break;
        }
        k += l0;
      } else {
        int h = 1ll * (f[k + l1] + z - f[k]) % z * invp[k] % z;
        if (h != h1) {
          flag = false;
          break;
        }
        k += l1;
      }
    }
    if (flag) {
      ans++;
    }
  }
  cout << ans << endl;
  return 0;
}

### 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;
const int p = 233;
const int z = 998244353;
int fp(int a, int x) {
  if (x == 0) {
    return 1;
  }
  if (x % 2 == 1) {
    return 1ll * a * fp(a, x - 1) % z;
  }
  int t = fp(a, x / 2);
  return 1ll * t * t % z;
}
int inv(int x) { return fp(x, z - 2); }
int main() {
  ios::sync_with_stdio(false);
  string s, t;
  cin >> s;
  cin >> t;
  if (s[0] == '1') {
    for (int i = 0; i < s.length(); i++) {
      s[i] = '0' + '1' - s[i];
    }
  }
  int c0 = 0;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '0') {
      c0++;
    }
  }
  int c1 = s.length() - c0;
  int s1;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '1') {
      s1 = i;
      break;
    }
  }
  int f[t.length() + 1];
  f[0] = 0;
  int q = 1;
  for (int i = 0; i < t.length(); i++) {
    int x = t[i] - 'a';
    f[i + 1] = (f[i] + 1ll * x * q) % z;
    q = 1ll * q * p % z;
  }
  int invp[t.length()];
  q = inv(p);
  invp[0] = 1;
  for (int i = 1; i < t.length(); i++) {
    invp[i] = 1ll * invp[i - 1] * q % z;
  }
  int ans = 0;
  for (int i = 1; i < t.length(); i++) {
    if (c0 * i >= t.length()) {
      break;
    }
    if ((t.length() - c0 * i) % c1 != 0) {
      continue;
    }
    int l0 = i, l1 = (t.length() - c0 * i) / c1;
    int h0 = f[l0];
    int h1 = 1ll * (f[s1 * l0 + l1] + z - f[s1 * l0]) % z * invp[s1 * l0] % z;
    if (l0 == l1 && h0 == h1) {
      continue;
    }
    int k = 0;
    bool flag = true;
    for (int j = 0; j < s.length(); j++) {
      if (s[j] == '0') {
        int h = 1ll * (f[k + l0] + z - f[k]) % z * invp[k] % z;
        if (h != h0) {
          flag = false;
          break;
        }
        k += l0;
      } else {
        int h = 1ll * (f[k + l1] + z - f[k]) % z * invp[k] % z;
        if (h != h1) {
          flag = false;
          break;
        }
        k += l1;
      }
    }
    if (flag) {
      ans++;
    }
  }
  cout << ans << endl;
  return 0;
}

```