#include <bits/stdc++.h>
using namespace std;
struct Str {
  vector<int64_t> h, d;
  const int64_t P = 31, MOD = 1e9 + 7;
  int64_t md(int64_t a) { return (a + MOD) % MOD; }
  Str() {}
  Str(string str) {
    h.resize(str.size() + 1);
    d.resize(str.size() + 1);
    d[0] = 1;
    for (int64_t i = 0; i < str.size(); ++i) {
      h[i + 1] = md(h[i] * P + str[i] - 'a' + 1);
      d[i + 1] = md(d[i] * P);
    }
  }
  int64_t hsh(int64_t l, int64_t r) {
    return md(h[r + 1] - md(h[l] * d[r - l + 1]));
  }
};
bool ch(int64_t &o, int64_t hsh) {
  if (o == -1) {
    o = hsh;
    return true;
  }
  return (hsh == o);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  string a, b;
  cin >> a >> b;
  int64_t A = 0, B = 0, ans = 0;
  for (int64_t i = 0; i < a.size(); ++i) {
    if (a[i] == '0')
      ++A;
    else
      ++B;
  }
  Str str = Str(b);
  for (int64_t i = 1; i <= b.size(); ++i) {
    if (i * A >= b.size()) break;
    int64_t j = b.size() - i * A;
    if (j % B) continue;
    j /= B;
    int64_t tec = 0;
    int64_t o1 = -1, o2 = -1;
    bool b = true;
    for (int64_t t = 0; t < a.size(); ++t) {
      if (a[t] == '0') {
        b &= ch(o1, str.hsh(tec, tec + i - 1));
        tec += i;
      } else {
        b &= ch(o2, str.hsh(tec, tec + j - 1));
        tec += j;
      }
    };
    if (b && o1 != o2) ++ans;
  }
  cout << ans;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 Str {
  vector<int64_t> h, d;
  const int64_t P = 31, MOD = 1e9 + 7;
  int64_t md(int64_t a) { return (a + MOD) % MOD; }
  Str() {}
  Str(string str) {
    h.resize(str.size() + 1);
    d.resize(str.size() + 1);
    d[0] = 1;
    for (int64_t i = 0; i < str.size(); ++i) {
      h[i + 1] = md(h[i] * P + str[i] - 'a' + 1);
      d[i + 1] = md(d[i] * P);
    }
  }
  int64_t hsh(int64_t l, int64_t r) {
    return md(h[r + 1] - md(h[l] * d[r - l + 1]));
  }
};
bool ch(int64_t &o, int64_t hsh) {
  if (o == -1) {
    o = hsh;
    return true;
  }
  return (hsh == o);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  string a, b;
  cin >> a >> b;
  int64_t A = 0, B = 0, ans = 0;
  for (int64_t i = 0; i < a.size(); ++i) {
    if (a[i] == '0')
      ++A;
    else
      ++B;
  }
  Str str = Str(b);
  for (int64_t i = 1; i <= b.size(); ++i) {
    if (i * A >= b.size()) break;
    int64_t j = b.size() - i * A;
    if (j % B) continue;
    j /= B;
    int64_t tec = 0;
    int64_t o1 = -1, o2 = -1;
    bool b = true;
    for (int64_t t = 0; t < a.size(); ++t) {
      if (a[t] == '0') {
        b &= ch(o1, str.hsh(tec, tec + i - 1));
        tec += i;
      } else {
        b &= ch(o2, str.hsh(tec, tec + j - 1));
        tec += j;
      }
    };
    if (b && o1 != o2) ++ans;
  }
  cout << ans;
}

```