#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
long long quick_pow(long long a, long long n, long long mod) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ans;
}
struct HS {
  long long has[N];
  long long pw[N], inv[N];
  const static long long p = 10007;
  const static long long mod = 998244353;
  void init(const char* s) {
    int n = strlen(s);
    has[0] = s[0];
    pw[0] = 1;
    for (int i = 1; i < n; i++) {
      pw[i] = pw[i - 1] * p % mod;
      has[i] = (has[i - 1] + s[i] * pw[i]) % mod;
    }
    inv[0] = 1;
    inv[1] = quick_pow(p, mod - 2, mod);
    for (int i = 2; i < n; i++) inv[i] = inv[i - 1] * inv[1] % mod;
  }
  long long getval(int l, int r) {
    if (l == 0) return has[r];
    return (has[r] - has[l - 1] + mod) * inv[l] % mod;
  }
} hs;
char s[N];
char t[N];
void solve() {
  scanf("%s%s", s, t);
  int n = strlen(s), m = strlen(t);
  if (s[0] == '1') {
    for (int i = 0; i < n; i++) {
      if (s[i] == '0')
        s[i] = '1';
      else
        s[i] = '0';
    }
  }
  int zero = 0, one = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '0')
      zero++;
    else
      one++;
  }
  hs.init(t);
  int ans = 0;
  for (int i = 1; zero * i < m; i++) {
    int left = m - i * zero;
    if (left % one) continue;
    int j = left / one;
    long long has0 = -1, has1 = -1;
    int l = 0;
    bool flag = 1;
    for (int k = 0; k < n; k++) {
      if (s[k] == '0') {
        if (has0 == -1)
          has0 = hs.getval(l, l + i - 1);
        else {
          if (has0 != hs.getval(l, l + i - 1)) {
            flag = 0;
            break;
          }
        }
        l += i;
      } else {
        if (has1 == -1)
          has1 = hs.getval(l, l + j - 1);
        else {
          if (has1 != hs.getval(l, l + j - 1)) {
            flag = 0;
            break;
          }
        }
        l += j;
      }
    }
    if (flag && has0 != has1) {
      ans++;
    }
  }
  cout << ans << endl;
}
int main() {
  solve();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 N = 1e6 + 10;
long long quick_pow(long long a, long long n, long long mod) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ans;
}
struct HS {
  long long has[N];
  long long pw[N], inv[N];
  const static long long p = 10007;
  const static long long mod = 998244353;
  void init(const char* s) {
    int n = strlen(s);
    has[0] = s[0];
    pw[0] = 1;
    for (int i = 1; i < n; i++) {
      pw[i] = pw[i - 1] * p % mod;
      has[i] = (has[i - 1] + s[i] * pw[i]) % mod;
    }
    inv[0] = 1;
    inv[1] = quick_pow(p, mod - 2, mod);
    for (int i = 2; i < n; i++) inv[i] = inv[i - 1] * inv[1] % mod;
  }
  long long getval(int l, int r) {
    if (l == 0) return has[r];
    return (has[r] - has[l - 1] + mod) * inv[l] % mod;
  }
} hs;
char s[N];
char t[N];
void solve() {
  scanf("%s%s", s, t);
  int n = strlen(s), m = strlen(t);
  if (s[0] == '1') {
    for (int i = 0; i < n; i++) {
      if (s[i] == '0')
        s[i] = '1';
      else
        s[i] = '0';
    }
  }
  int zero = 0, one = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '0')
      zero++;
    else
      one++;
  }
  hs.init(t);
  int ans = 0;
  for (int i = 1; zero * i < m; i++) {
    int left = m - i * zero;
    if (left % one) continue;
    int j = left / one;
    long long has0 = -1, has1 = -1;
    int l = 0;
    bool flag = 1;
    for (int k = 0; k < n; k++) {
      if (s[k] == '0') {
        if (has0 == -1)
          has0 = hs.getval(l, l + i - 1);
        else {
          if (has0 != hs.getval(l, l + i - 1)) {
            flag = 0;
            break;
          }
        }
        l += i;
      } else {
        if (has1 == -1)
          has1 = hs.getval(l, l + j - 1);
        else {
          if (has1 != hs.getval(l, l + j - 1)) {
            flag = 0;
            break;
          }
        }
        l += j;
      }
    }
    if (flag && has0 != has1) {
      ans++;
    }
  }
  cout << ans << endl;
}
int main() {
  solve();
  return 0;
}

```