#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7;
const int K = 41;
const int MAX = 1e6 + 5;
int h[MAX], pw[MAX], len[2], cnt[2], g[2], f[2];
int get(int l, int r) {
  return (h[r] - 1ll * h[l - 1] * pw[r - l + 1] % P + P) % P;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s, t;
  cin >> s >> t;
  int n = s.size(), m = t.size();
  s = ' ' + s;
  t = ' ' + t;
  for (int i = 1; i <= n; i++) cnt[s[i] - '0']++;
  pw[0] = 1;
  for (int i = 1; i <= m; i++) {
    pw[i] = (1ll * pw[i - 1] * K) % P;
    h[i] = (1ll * h[i - 1] * K + t[i] - 'a' + 1) % P;
  }
  int ans = 0;
  for (int i = 1;; i++) {
    f[0] = f[1] = 1;
    if (cnt[0] * i >= m) break;
    if ((m - cnt[0] * i) % cnt[1]) continue;
    len[0] = i;
    len[1] = (m - cnt[0] * i) / cnt[1];
    for (int l = 1, k = 1; k <= n; k++) {
      int cur = s[k] - '0';
      if (f[cur])
        g[cur] = get(l, l + len[cur] - 1), f[cur] = 0;
      else if (g[cur] != get(l, l + len[cur] - 1))
        break;
      if (!f[0] && !f[1] && g[0] == g[1]) break;
      if (k == n) ans++;
      l += len[cur];
    }
  }
  cout << ans;
  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 P = 1e9 + 7;
const int K = 41;
const int MAX = 1e6 + 5;
int h[MAX], pw[MAX], len[2], cnt[2], g[2], f[2];
int get(int l, int r) {
  return (h[r] - 1ll * h[l - 1] * pw[r - l + 1] % P + P) % P;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s, t;
  cin >> s >> t;
  int n = s.size(), m = t.size();
  s = ' ' + s;
  t = ' ' + t;
  for (int i = 1; i <= n; i++) cnt[s[i] - '0']++;
  pw[0] = 1;
  for (int i = 1; i <= m; i++) {
    pw[i] = (1ll * pw[i - 1] * K) % P;
    h[i] = (1ll * h[i - 1] * K + t[i] - 'a' + 1) % P;
  }
  int ans = 0;
  for (int i = 1;; i++) {
    f[0] = f[1] = 1;
    if (cnt[0] * i >= m) break;
    if ((m - cnt[0] * i) % cnt[1]) continue;
    len[0] = i;
    len[1] = (m - cnt[0] * i) / cnt[1];
    for (int l = 1, k = 1; k <= n; k++) {
      int cur = s[k] - '0';
      if (f[cur])
        g[cur] = get(l, l + len[cur] - 1), f[cur] = 0;
      else if (g[cur] != get(l, l + len[cur] - 1))
        break;
      if (!f[0] && !f[1] && g[0] == g[1]) break;
      if (k == n) ans++;
      l += len[cur];
    }
  }
  cout << ans;
  return 0;
}

```