#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:60777216")
using namespace std;
string s, t;
long long h1[1 << 20];
long long m1[1 << 20];
long long h2[1 << 20];
long long m2[1 << 20];
long long getH(long long* h, long long* m, int pos, int len) {
  return h[pos + len - 1] - (pos - 1 >= 0 ? h[pos - 1] : 0) * m[len];
}
pair<long long, long long> getH(int pos, int len) {
  return pair<long long, long long>(getH(h1, m1, pos, len),
                                    getH(h2, m2, pos, len));
}
bool good(int l0, int l1) {
  pair<long long, long long> h0, h1;
  bool hh0 = false, hh1 = false;
  int pos = 0;
  for (int i = (0); i < (s.size()); i++) {
    if (s[i] == '0') {
      pair<long long, long long> x = getH(pos, l0);
      if (!hh0)
        h0 = x, hh0 = 1;
      else if (h0 != x)
        return 0;
      pos += l0;
    } else {
      pair<long long, long long> x = getH(pos, l1);
      if (!hh1)
        h1 = x, hh1 = 1;
      else if (h1 != x)
        return 0;
      pos += l1;
    }
  }
  if (l0 == l1 && h0 == h1) return 0;
  return 1;
}
int main() {
  cin >> s >> t;
  long long n0 = 0, n1 = 0;
  for (int i = (0); i < (s.size()); i++)
    if (s[i] == '0')
      n0++;
    else
      n1++;
  m1[0] = m2[0] = 1;
  long long md = rand() % 10000 + 100000011;
  for (int i = (1); i < (1 << 20); i++) {
    m1[i] = m1[i - 1] * md;
    m2[i] = m2[i - 1] * 1000000007;
  }
  for (int i = (0); i < (t.size()); i++) {
    h1[i] = (i - 1 >= 0 ? h1[i - 1] : 0) * m1[1] + t[i] - 'a' + 1;
    h2[i] = (i - 1 >= 0 ? h2[i - 1] : 0) * m2[1] + t[i] - 'a' + 1;
  }
  int res = 0;
  for (int l0 = (1); l0 < (t.size() + 1); l0++)
    if (n0 * l0 < t.size()) {
      long long rem = t.size() - n0 * l0;
      if (rem % n1 == 0) {
        long long l1 = rem / n1;
        if (good(l0, l1)) {
          res++;
        }
      }
    }
  cout << res << endl;
  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>
#pragma comment(linker, "/STACK:60777216")
using namespace std;
string s, t;
long long h1[1 << 20];
long long m1[1 << 20];
long long h2[1 << 20];
long long m2[1 << 20];
long long getH(long long* h, long long* m, int pos, int len) {
  return h[pos + len - 1] - (pos - 1 >= 0 ? h[pos - 1] : 0) * m[len];
}
pair<long long, long long> getH(int pos, int len) {
  return pair<long long, long long>(getH(h1, m1, pos, len),
                                    getH(h2, m2, pos, len));
}
bool good(int l0, int l1) {
  pair<long long, long long> h0, h1;
  bool hh0 = false, hh1 = false;
  int pos = 0;
  for (int i = (0); i < (s.size()); i++) {
    if (s[i] == '0') {
      pair<long long, long long> x = getH(pos, l0);
      if (!hh0)
        h0 = x, hh0 = 1;
      else if (h0 != x)
        return 0;
      pos += l0;
    } else {
      pair<long long, long long> x = getH(pos, l1);
      if (!hh1)
        h1 = x, hh1 = 1;
      else if (h1 != x)
        return 0;
      pos += l1;
    }
  }
  if (l0 == l1 && h0 == h1) return 0;
  return 1;
}
int main() {
  cin >> s >> t;
  long long n0 = 0, n1 = 0;
  for (int i = (0); i < (s.size()); i++)
    if (s[i] == '0')
      n0++;
    else
      n1++;
  m1[0] = m2[0] = 1;
  long long md = rand() % 10000 + 100000011;
  for (int i = (1); i < (1 << 20); i++) {
    m1[i] = m1[i - 1] * md;
    m2[i] = m2[i - 1] * 1000000007;
  }
  for (int i = (0); i < (t.size()); i++) {
    h1[i] = (i - 1 >= 0 ? h1[i - 1] : 0) * m1[1] + t[i] - 'a' + 1;
    h2[i] = (i - 1 >= 0 ? h2[i - 1] : 0) * m2[1] + t[i] - 'a' + 1;
  }
  int res = 0;
  for (int l0 = (1); l0 < (t.size() + 1); l0++)
    if (n0 * l0 < t.size()) {
      long long rem = t.size() - n0 * l0;
      if (rem % n1 == 0) {
        long long l1 = rem / n1;
        if (good(l0, l1)) {
          res++;
        }
      }
    }
  cout << res << endl;
  return 0;
}

```