#include <bits/stdc++.h>
using namespace std;
class string_hash {
 public:
  struct hash {
    int MOD1 = 1e9 + 7, MOD2 = 1e9 + 9;
    long long l, r;
    long long norm(long long x, int MOD) {
      x %= MOD;
      if (x < 0) x += MOD;
      return x;
    }
    hash(long long l, long long r) : l(norm(l, MOD1)), r(norm(r, MOD2)) {}
    hash(long long x = 0) : l(x), r(x) {}
    hash operator+(const hash& a) const { return hash(l + a.l, r + a.r); }
    hash operator-(const hash& a) const { return hash(l - a.l, r - a.r); }
    hash operator*(const hash& a) const { return hash(l * a.l, r * a.r); }
    bool operator==(const hash& a) const { return l == a.l and r == a.r; }
  };
  const int P = 31;
  vector<hash> p, h;
  int n;
  string_hash(const string& s) : n(s.size()) {
    p.resize(n + 1);
    h.resize(n + 1);
    p[0] = 1;
    for (int i = 1; i <= n; ++i) {
      p[i] = p[i - 1] * P;
      h[i] = h[i - 1] + p[i - 1] * (s[i - 1] - 'a');
    }
  }
  bool is_equal(int i, int j, int len) {
    if (i > j) {
      swap(i, j);
    }
    return ((h[i + len] - h[i]) * p[j - i]) == (h[j + len] - h[j]);
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s, t;
  cin >> s >> t;
  int n = s.size(), m = t.size();
  string_hash h(t);
  int a = count(s.begin(), s.end(), '1'), b = n - a;
  int ans = 0;
  for (int x = 1; x * a < m; ++x) {
    if ((m - x * a) % b) {
      continue;
    }
    int y = (m - x * a) / b;
    int ptr0 = -1, ptr1 = -1, ptr = 0;
    bool ok = true;
    for (char c : s) {
      if (c == '0') {
        if (ptr0 == -1) {
          ptr0 = ptr;
        } else {
          ok &= h.is_equal(ptr0, ptr, y);
        }
        ptr += y;
      } else {
        if (ptr1 == -1) {
          ptr1 = ptr;
        } else {
          ok &= h.is_equal(ptr1, ptr, x);
        }
        ptr += x;
      }
    }
    if (x == y) {
      ok &= !(h.is_equal(ptr0, ptr1, x));
    }
    ans += ok;
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in cpp 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;
class string_hash {
 public:
  struct hash {
    int MOD1 = 1e9 + 7, MOD2 = 1e9 + 9;
    long long l, r;
    long long norm(long long x, int MOD) {
      x %= MOD;
      if (x < 0) x += MOD;
      return x;
    }
    hash(long long l, long long r) : l(norm(l, MOD1)), r(norm(r, MOD2)) {}
    hash(long long x = 0) : l(x), r(x) {}
    hash operator+(const hash& a) const { return hash(l + a.l, r + a.r); }
    hash operator-(const hash& a) const { return hash(l - a.l, r - a.r); }
    hash operator*(const hash& a) const { return hash(l * a.l, r * a.r); }
    bool operator==(const hash& a) const { return l == a.l and r == a.r; }
  };
  const int P = 31;
  vector<hash> p, h;
  int n;
  string_hash(const string& s) : n(s.size()) {
    p.resize(n + 1);
    h.resize(n + 1);
    p[0] = 1;
    for (int i = 1; i <= n; ++i) {
      p[i] = p[i - 1] * P;
      h[i] = h[i - 1] + p[i - 1] * (s[i - 1] - 'a');
    }
  }
  bool is_equal(int i, int j, int len) {
    if (i > j) {
      swap(i, j);
    }
    return ((h[i + len] - h[i]) * p[j - i]) == (h[j + len] - h[j]);
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s, t;
  cin >> s >> t;
  int n = s.size(), m = t.size();
  string_hash h(t);
  int a = count(s.begin(), s.end(), '1'), b = n - a;
  int ans = 0;
  for (int x = 1; x * a < m; ++x) {
    if ((m - x * a) % b) {
      continue;
    }
    int y = (m - x * a) / b;
    int ptr0 = -1, ptr1 = -1, ptr = 0;
    bool ok = true;
    for (char c : s) {
      if (c == '0') {
        if (ptr0 == -1) {
          ptr0 = ptr;
        } else {
          ok &= h.is_equal(ptr0, ptr, y);
        }
        ptr += y;
      } else {
        if (ptr1 == -1) {
          ptr1 = ptr;
        } else {
          ok &= h.is_equal(ptr1, ptr, x);
        }
        ptr += x;
      }
    }
    if (x == y) {
      ok &= !(h.is_equal(ptr0, ptr1, x));
    }
    ans += ok;
  }
  cout << ans;
  return 0;
}

```