#include <bits/stdc++.h>
using namespace std;
struct strLine {
  long double a, b, c;
};
const double pi = 3.141592653589793;
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long max(long long a, long long b) { return (a >= b) ? a : b; }
long long min(long long a, long long b) { return (a <= b) ? a : b; }
const pair<long long, long long> mod =
    pair<long long, long long>(1000000009, 1007050321);
const pair<long long, long long> base = pair<long long, long long>(337, 337);
long n, m, c0, c1;
string s, t;
pair<long long, long long> h[1000006], pw[1000006];
pair<long long, long long> operator%(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return pair<long long, long long>(a.first % b.first, a.second % b.second);
}
pair<long long, long long> operator*(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return pair<long long, long long>(a.first * b.first, a.second * b.second) %
         mod;
}
pair<long long, long long> operator+(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return pair<long long, long long>(a.first + b.first, a.second + b.second) %
         mod;
}
pair<long long, long long> operator-(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return (pair<long long, long long>(a.first - b.first, b.second - b.second) %
              mod +
          mod) %
         mod;
}
pair<long long, long long> get_hash(long l, long len) {
  return h[l + len - 1] - h[l - 1] * pw[len];
}
long chk(long len) {
  if (1LL * len * c0 >= m) return 0;
  if ((m - len * c0) % c1 != 0) return 0;
  long len1 = (m - len * c0) / c1;
  pair<long long, long long> key0 = get_hash(1, len);
  pair<long long, long long> key1 = pair<long long, long long>(-1, -1);
  long ptr = 1;
  for (int i = 0; i <= n - 1; i++) {
    if (s[i] == '0') {
      if (get_hash(ptr, len) != key0) return 0;
      ptr += len;
    } else {
      if (key1.first == -1)
        key1 = get_hash(ptr, len1);
      else if (get_hash(ptr, len1) != key1)
        return 0;
      ptr += len1;
    }
  }
  if (key0 == key1) return 0;
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> s >> t;
  n = s.size();
  m = t.size();
  if (s[0] == '1')
    for (int i = 0; i <= n - 1; i++)
      if (s[i] == '1')
        s[i] = '0';
      else
        s[i] = '1';
  for (int i = 0; i < n; i++)
    if (s[i] == '0')
      c0++;
    else
      c1++;
  for (int i = 1; i <= m; i++)
    h[i] = h[i - 1] * base + pair<long long, long long>(t[i - 1], t[i - 1]);
  pw[0] = pair<long long, long long>(1, 1);
  for (int i = 1; i <= m; i++) pw[i] = pw[i - 1] * base;
  long ans = 0;
  for (int i = 1; i <= m - 1; i++) ans += chk(i);
  cout << ans;
  return 0;
}

### Prompt

Please formulate a Cpp solution to 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;
struct strLine {
  long double a, b, c;
};
const double pi = 3.141592653589793;
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long max(long long a, long long b) { return (a >= b) ? a : b; }
long long min(long long a, long long b) { return (a <= b) ? a : b; }
const pair<long long, long long> mod =
    pair<long long, long long>(1000000009, 1007050321);
const pair<long long, long long> base = pair<long long, long long>(337, 337);
long n, m, c0, c1;
string s, t;
pair<long long, long long> h[1000006], pw[1000006];
pair<long long, long long> operator%(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return pair<long long, long long>(a.first % b.first, a.second % b.second);
}
pair<long long, long long> operator*(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return pair<long long, long long>(a.first * b.first, a.second * b.second) %
         mod;
}
pair<long long, long long> operator+(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return pair<long long, long long>(a.first + b.first, a.second + b.second) %
         mod;
}
pair<long long, long long> operator-(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return (pair<long long, long long>(a.first - b.first, b.second - b.second) %
              mod +
          mod) %
         mod;
}
pair<long long, long long> get_hash(long l, long len) {
  return h[l + len - 1] - h[l - 1] * pw[len];
}
long chk(long len) {
  if (1LL * len * c0 >= m) return 0;
  if ((m - len * c0) % c1 != 0) return 0;
  long len1 = (m - len * c0) / c1;
  pair<long long, long long> key0 = get_hash(1, len);
  pair<long long, long long> key1 = pair<long long, long long>(-1, -1);
  long ptr = 1;
  for (int i = 0; i <= n - 1; i++) {
    if (s[i] == '0') {
      if (get_hash(ptr, len) != key0) return 0;
      ptr += len;
    } else {
      if (key1.first == -1)
        key1 = get_hash(ptr, len1);
      else if (get_hash(ptr, len1) != key1)
        return 0;
      ptr += len1;
    }
  }
  if (key0 == key1) return 0;
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> s >> t;
  n = s.size();
  m = t.size();
  if (s[0] == '1')
    for (int i = 0; i <= n - 1; i++)
      if (s[i] == '1')
        s[i] = '0';
      else
        s[i] = '1';
  for (int i = 0; i < n; i++)
    if (s[i] == '0')
      c0++;
    else
      c1++;
  for (int i = 1; i <= m; i++)
    h[i] = h[i - 1] * base + pair<long long, long long>(t[i - 1], t[i - 1]);
  pw[0] = pair<long long, long long>(1, 1);
  for (int i = 1; i <= m; i++) pw[i] = pw[i - 1] * base;
  long ans = 0;
  for (int i = 1; i <= m - 1; i++) ans += chk(i);
  cout << ans;
  return 0;
}

```