#include <bits/stdc++.h>
const long long M = 10000019;
const long long INV = 4814824;
using namespace std;
long long n, m, i, j, k, h[1000005], inv27[1000005], p27[1000005];
string s, pat;
long long geth(long l, long r) {
  return (h[r] - h[l - 1] + M) % M * inv27[l - 1] % M;
}
int main() {
  cin >> s >> pat;
  n = s.size(), m = pat.size();
  long c[2] = {0, 0}, cnt = 0;
  for (i = 0; i < n; i++) c[s[i] - '0']++;
  inv27[0] = p27[0] = 1;
  for (i = 0, k = 1; i < m; i++, k = k * 27 % M)
    h[i + 1] = (h[i] + k * (pat[i] - 'a' + 1)) % M,
          p27[i + 1] = p27[i] * 27 % M, inv27[i + 1] = inv27[i] * INV % M;
  for (i = 1; c[0] * i < m; i++) {
    if ((m - c[0] * i) % c[1] != 0) continue;
    long long l0 = i, l1 = (m - c[0] * i) / c[1], tmp = 0, h0, h1;
    bool fl = 1, f1 = 0, f0 = 0;
    for (j = k = 0; j < n; j++) {
      if (s[j] == '0') {
        if (!f0) {
          f0 = 1;
          h0 = geth(k + 1, k + l0);
        }
        tmp = (tmp + h0 * p27[k] % M) % M;
        k += l0;
      } else {
        if (!f1) {
          f1 = 1;
          h1 = geth(k + 1, k + l1);
        }
        tmp = (tmp + h1 * p27[k] % M) % M;
        k += l1;
      }
      if (tmp != h[k]) {
        fl = 0;
        break;
      }
    }
    if (fl && h0 != h1) cnt++;
  }
  cout << cnt;
  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>
const long long M = 10000019;
const long long INV = 4814824;
using namespace std;
long long n, m, i, j, k, h[1000005], inv27[1000005], p27[1000005];
string s, pat;
long long geth(long l, long r) {
  return (h[r] - h[l - 1] + M) % M * inv27[l - 1] % M;
}
int main() {
  cin >> s >> pat;
  n = s.size(), m = pat.size();
  long c[2] = {0, 0}, cnt = 0;
  for (i = 0; i < n; i++) c[s[i] - '0']++;
  inv27[0] = p27[0] = 1;
  for (i = 0, k = 1; i < m; i++, k = k * 27 % M)
    h[i + 1] = (h[i] + k * (pat[i] - 'a' + 1)) % M,
          p27[i + 1] = p27[i] * 27 % M, inv27[i + 1] = inv27[i] * INV % M;
  for (i = 1; c[0] * i < m; i++) {
    if ((m - c[0] * i) % c[1] != 0) continue;
    long long l0 = i, l1 = (m - c[0] * i) / c[1], tmp = 0, h0, h1;
    bool fl = 1, f1 = 0, f0 = 0;
    for (j = k = 0; j < n; j++) {
      if (s[j] == '0') {
        if (!f0) {
          f0 = 1;
          h0 = geth(k + 1, k + l0);
        }
        tmp = (tmp + h0 * p27[k] % M) % M;
        k += l0;
      } else {
        if (!f1) {
          f1 = 1;
          h1 = geth(k + 1, k + l1);
        }
        tmp = (tmp + h1 * p27[k] % M) % M;
        k += l1;
      }
      if (tmp != h[k]) {
        fl = 0;
        break;
      }
    }
    if (fl && h0 != h1) cnt++;
  }
  cout << cnt;
  return 0;
}

```