#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
const long long base = 233;
const long long mod = 1e9 + 7;
char s[100050], t[1000050];
int n, m;
int cnt[2];
long long h[1000050], b[1000050], f[2], e[2], g[2], ans;
inline long long GetHash(int l, int r) {
  return (h[r] - h[l - 1] * b[r - l + 1] % mod + mod) % mod;
}
int main() {
  cin >> (s + 1) >> (t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (int i = 1; i <= n; i++) cnt[s[i] - '0']++;
  b[0] = 1;
  for (int i = 1; i <= m; i++) {
    h[i] = (h[i - 1] * base % mod + (t[i] - 'a' + 1)) % mod;
    b[i] = b[i - 1] * base % mod;
  }
  for (int i = 1; cnt[0] * i < m; i++) {
    f[0] = f[1] = 1;
    if ((m - cnt[0] * i) % cnt[1]) continue;
    e[0] = i;
    e[1] = ((m - cnt[0] * i) / cnt[1]);
    for (int l = 1, k = 1; k <= n; k++) {
      int res = s[k] - '0';
      if (f[res]) {
        g[res] = GetHash(l, l + e[res] - 1);
        f[res] = 0;
      } else {
        if (g[res] != GetHash(l, l + e[res] - 1)) break;
      }
      if (!f[0] && !f[1] && (g[0] == g[1])) break;
      if (k == n) ans++;
      l += e[res];
    }
  }
  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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
const long long base = 233;
const long long mod = 1e9 + 7;
char s[100050], t[1000050];
int n, m;
int cnt[2];
long long h[1000050], b[1000050], f[2], e[2], g[2], ans;
inline long long GetHash(int l, int r) {
  return (h[r] - h[l - 1] * b[r - l + 1] % mod + mod) % mod;
}
int main() {
  cin >> (s + 1) >> (t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (int i = 1; i <= n; i++) cnt[s[i] - '0']++;
  b[0] = 1;
  for (int i = 1; i <= m; i++) {
    h[i] = (h[i - 1] * base % mod + (t[i] - 'a' + 1)) % mod;
    b[i] = b[i - 1] * base % mod;
  }
  for (int i = 1; cnt[0] * i < m; i++) {
    f[0] = f[1] = 1;
    if ((m - cnt[0] * i) % cnt[1]) continue;
    e[0] = i;
    e[1] = ((m - cnt[0] * i) / cnt[1]);
    for (int l = 1, k = 1; k <= n; k++) {
      int res = s[k] - '0';
      if (f[res]) {
        g[res] = GetHash(l, l + e[res] - 1);
        f[res] = 0;
      } else {
        if (g[res] != GetHash(l, l + e[res] - 1)) break;
      }
      if (!f[0] && !f[1] && (g[0] == g[1])) break;
      if (k == n) ans++;
      l += e[res];
    }
  }
  cout << ans << endl;
  return 0;
}

```