#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 5;
long long MOD[] = {1000000000 + 7, 1000000000 + 9};
int m, n;
int z, o;
string s, t;
long long P[] = {29, 31};
long long hasj[2][N], power[2][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s >> t;
  m = s.size();
  n = t.size();
  if (s[0] != '0')
    for (int i = 0; i < m; i++) {
      s[i] = (s[i] == '0') ? '1' : '0';
    }
  for (int i = 0; i < m; i++) {
    if (s[i] == '0')
      z++;
    else
      o++;
  }
  power[0][0] = power[1][0] = 1;
  for (int i = 0; i < n; i++) {
    hasj[0][i + 1] = ((hasj[0][i] * P[0]) + (t[i] - 'a' + 1)) % MOD[0];
    power[0][i + 1] = (power[0][i] * P[0]) % MOD[0];
    hasj[1][i + 1] = ((hasj[1][i] * P[1]) + (t[i] - 'a' + 1)) % MOD[1];
    power[1][i + 1] = (power[1][i] * P[1]) % MOD[1];
  }
  int ans = 0;
  for (int i = 1; i * z < n; i++) {
    int x = n - z * i;
    if (x % o) continue;
    int j = x / o;
    int ind = 0, gone = 0;
    long long h0[] = {-1, -1}, h1[] = {-1, -1};
    for (int k = 0; k < m; k++) {
      if (s[k] == '0') {
        long long x[] = {(hasj[0][ind + i] +
                          (MOD[0] - (hasj[0][ind] * power[0][i]) % MOD[0])) %
                             MOD[0],
                         (hasj[1][ind + i] +
                          (MOD[1] - (hasj[1][ind] * power[1][i]) % MOD[1])) %
                             MOD[1]};
        if (h0[0] == -1) {
          h0[0] = x[0];
          h0[1] = x[1];
        } else if ((h0[0] != x[0]) or (h0[1] != x[1])) {
          gone = 1;
          break;
        }
        ind += i;
      } else
        ind += j;
    }
    if (gone) continue;
    ind = gone = 0;
    for (int k = 0; k < m; k++) {
      if (s[k] == '1') {
        long long x[] = {(hasj[0][ind + j] +
                          (MOD[0] - (hasj[0][ind] * power[0][j]) % MOD[0])) %
                             MOD[0],
                         (hasj[1][ind + j] +
                          (MOD[1] - (hasj[1][ind] * power[1][j]) % MOD[1])) %
                             MOD[1]};
        if (h1[0] == -1) {
          h1[0] = x[0];
          h1[1] = x[1];
        } else if ((h1[0] != x[0]) or (h1[1] != x[1])) {
          gone = 1;
          break;
        }
        ind += j;
      } else
        ind += i;
    }
    if (gone) continue;
    if (h0[0] == h1[0] and h0[1] == h1[1]) continue;
    ans++;
  }
  printf("%d\n", ans);
}

### Prompt

Generate 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;
const long long N = 1e6 + 5;
long long MOD[] = {1000000000 + 7, 1000000000 + 9};
int m, n;
int z, o;
string s, t;
long long P[] = {29, 31};
long long hasj[2][N], power[2][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s >> t;
  m = s.size();
  n = t.size();
  if (s[0] != '0')
    for (int i = 0; i < m; i++) {
      s[i] = (s[i] == '0') ? '1' : '0';
    }
  for (int i = 0; i < m; i++) {
    if (s[i] == '0')
      z++;
    else
      o++;
  }
  power[0][0] = power[1][0] = 1;
  for (int i = 0; i < n; i++) {
    hasj[0][i + 1] = ((hasj[0][i] * P[0]) + (t[i] - 'a' + 1)) % MOD[0];
    power[0][i + 1] = (power[0][i] * P[0]) % MOD[0];
    hasj[1][i + 1] = ((hasj[1][i] * P[1]) + (t[i] - 'a' + 1)) % MOD[1];
    power[1][i + 1] = (power[1][i] * P[1]) % MOD[1];
  }
  int ans = 0;
  for (int i = 1; i * z < n; i++) {
    int x = n - z * i;
    if (x % o) continue;
    int j = x / o;
    int ind = 0, gone = 0;
    long long h0[] = {-1, -1}, h1[] = {-1, -1};
    for (int k = 0; k < m; k++) {
      if (s[k] == '0') {
        long long x[] = {(hasj[0][ind + i] +
                          (MOD[0] - (hasj[0][ind] * power[0][i]) % MOD[0])) %
                             MOD[0],
                         (hasj[1][ind + i] +
                          (MOD[1] - (hasj[1][ind] * power[1][i]) % MOD[1])) %
                             MOD[1]};
        if (h0[0] == -1) {
          h0[0] = x[0];
          h0[1] = x[1];
        } else if ((h0[0] != x[0]) or (h0[1] != x[1])) {
          gone = 1;
          break;
        }
        ind += i;
      } else
        ind += j;
    }
    if (gone) continue;
    ind = gone = 0;
    for (int k = 0; k < m; k++) {
      if (s[k] == '1') {
        long long x[] = {(hasj[0][ind + j] +
                          (MOD[0] - (hasj[0][ind] * power[0][j]) % MOD[0])) %
                             MOD[0],
                         (hasj[1][ind + j] +
                          (MOD[1] - (hasj[1][ind] * power[1][j]) % MOD[1])) %
                             MOD[1]};
        if (h1[0] == -1) {
          h1[0] = x[0];
          h1[1] = x[1];
        } else if ((h1[0] != x[0]) or (h1[1] != x[1])) {
          gone = 1;
          break;
        }
        ind += j;
      } else
        ind += i;
    }
    if (gone) continue;
    if (h0[0] == h1[0] and h0[1] == h1[1]) continue;
    ans++;
  }
  printf("%d\n", ans);
}

```