#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool minn(T &A, T B) {
  return A > B ? (A = B, 1) : 0;
}
template <class T>
inline bool maxx(T &A, T B) {
  return A < B ? (A = B, 1) : 0;
}
const long long mod = 1e9 + 7;
const long long base = 9973;
string s, t;
int n, cnt0, cnt1, cnt, res;
long long hsh[1000005], pw[1000005];
int get(int l, int r) {
  return (hsh[r] - hsh[l - 1] * pw[r - l + 1] + mod * mod) % mod;
}
bool ck(int dem0, int dem1) {
  int res1 = 0, res0 = 0, i = 1;
  for (char c : s) {
    if (c == '0') {
      if (res0 == 0)
        res0 = get(i, i + dem0 - 1);
      else if (res0 != get(i, i + dem0 - 1))
        return 0;
      i += dem0;
    } else {
      if (res1 == 0)
        res1 = get(i, i + dem1 - 1);
      else if (res1 != get(i, i + dem1 - 1))
        return 0;
      i += dem1;
    }
  }
  return res1 != res0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s >> t;
  n = (int)((t).size());
  t = ' ' + t;
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) {
    pw[i] = (pw[i - 1] * base) % mod;
    hsh[i] = (hsh[i - 1] * base + t[i] - 'a' + 1) % mod;
  }
  for (char c : s) {
    if (c == '0')
      cnt0++;
    else
      cnt1++;
  }
  for (int i = 1; i * cnt0 + cnt1 <= n; ++i) {
    cnt = n - i * cnt0;
    ;
    if (cnt % cnt1) continue;
    if (ck(i, cnt / cnt1)) res++;
  }
  cout << res;
}

### 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;
template <class T>
inline bool minn(T &A, T B) {
  return A > B ? (A = B, 1) : 0;
}
template <class T>
inline bool maxx(T &A, T B) {
  return A < B ? (A = B, 1) : 0;
}
const long long mod = 1e9 + 7;
const long long base = 9973;
string s, t;
int n, cnt0, cnt1, cnt, res;
long long hsh[1000005], pw[1000005];
int get(int l, int r) {
  return (hsh[r] - hsh[l - 1] * pw[r - l + 1] + mod * mod) % mod;
}
bool ck(int dem0, int dem1) {
  int res1 = 0, res0 = 0, i = 1;
  for (char c : s) {
    if (c == '0') {
      if (res0 == 0)
        res0 = get(i, i + dem0 - 1);
      else if (res0 != get(i, i + dem0 - 1))
        return 0;
      i += dem0;
    } else {
      if (res1 == 0)
        res1 = get(i, i + dem1 - 1);
      else if (res1 != get(i, i + dem1 - 1))
        return 0;
      i += dem1;
    }
  }
  return res1 != res0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s >> t;
  n = (int)((t).size());
  t = ' ' + t;
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) {
    pw[i] = (pw[i - 1] * base) % mod;
    hsh[i] = (hsh[i - 1] * base + t[i] - 'a' + 1) % mod;
  }
  for (char c : s) {
    if (c == '0')
      cnt0++;
    else
      cnt1++;
  }
  for (int i = 1; i * cnt0 + cnt1 <= n; ++i) {
    cnt = n - i * cnt0;
    ;
    if (cnt % cnt1) continue;
    if (ck(i, cnt / cnt1)) res++;
  }
  cout << res;
}

```