#include <bits/stdc++.h>
using namespace std;
string s, t;
long long pre[1100100], p = 31, cur;
long long expo(long long a, long long b) {
  long long res = 1;
  for (int i = (30); i >= (0); i--) {
    res = (res * res) % 1000000007;
    if ((1 << i) & b) res = (res * a) % 1000000007;
  }
  return res;
}
bool eq(int i1, int i2, int len) {
  long long hash1 = (pre[i1 + len] - pre[i1]) * expo(p, i2 - i1);
  long long hash2 = pre[i2 + len] - pre[i2];
  return (hash2 - hash1) % 1000000007 == 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> s >> t;
  cur = 1;
  for (int i = (1); i <= (t.size()); i++) {
    pre[i] = (t[i - 1] * cur + pre[i - 1]) % 1000000007;
    cur = (cur * p) % 1000000007;
  }
  int c0 = 0;
  for (auto c : s)
    if (c == '0') c0++;
  int c1 = s.size() - c0;
  if (c0 < c1) {
    for (int i = (0); i <= ((int)(s.size()) - 1); i++) s[i] = '0' + '1' - s[i];
    swap(c0, c1);
  }
  int ans = 0;
  for (int l0 = 1; l0 * c0 < t.size(); l0++)
    if ((t.size() - l0 * c0) % c1 == 0) {
      int l1 = (t.size() - l0 * c0) / c1;
      int id = 0, f0 = -1, f1 = -1;
      bool flag = false;
      for (auto c : s) {
        if (c == '0') {
          if (f0 == -1)
            f0 = id;
          else if (!eq(f0, id, l0))
            flag = true;
          id += l0;
        } else {
          if (f1 == -1)
            f1 = id;
          else if (!eq(f1, id, l1))
            flag = true;
          id += l1;
        }
      }
      if (l0 == l1 && eq(min(f0, f1), max(f0, f1), l0)) continue;
      if (!flag) ans++;
    }
  cout << ans;
}

### Prompt

Develop a solution in CPP 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;
string s, t;
long long pre[1100100], p = 31, cur;
long long expo(long long a, long long b) {
  long long res = 1;
  for (int i = (30); i >= (0); i--) {
    res = (res * res) % 1000000007;
    if ((1 << i) & b) res = (res * a) % 1000000007;
  }
  return res;
}
bool eq(int i1, int i2, int len) {
  long long hash1 = (pre[i1 + len] - pre[i1]) * expo(p, i2 - i1);
  long long hash2 = pre[i2 + len] - pre[i2];
  return (hash2 - hash1) % 1000000007 == 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> s >> t;
  cur = 1;
  for (int i = (1); i <= (t.size()); i++) {
    pre[i] = (t[i - 1] * cur + pre[i - 1]) % 1000000007;
    cur = (cur * p) % 1000000007;
  }
  int c0 = 0;
  for (auto c : s)
    if (c == '0') c0++;
  int c1 = s.size() - c0;
  if (c0 < c1) {
    for (int i = (0); i <= ((int)(s.size()) - 1); i++) s[i] = '0' + '1' - s[i];
    swap(c0, c1);
  }
  int ans = 0;
  for (int l0 = 1; l0 * c0 < t.size(); l0++)
    if ((t.size() - l0 * c0) % c1 == 0) {
      int l1 = (t.size() - l0 * c0) / c1;
      int id = 0, f0 = -1, f1 = -1;
      bool flag = false;
      for (auto c : s) {
        if (c == '0') {
          if (f0 == -1)
            f0 = id;
          else if (!eq(f0, id, l0))
            flag = true;
          id += l0;
        } else {
          if (f1 == -1)
            f1 = id;
          else if (!eq(f1, id, l1))
            flag = true;
          id += l1;
        }
      }
      if (l0 == l1 && eq(min(f0, f1), max(f0, f1), l0)) continue;
      if (!flag) ans++;
    }
  cout << ans;
}

```