#include <bits/stdc++.h>
using namespace std;
string s, t;
long long f[1000005], hashh[1000005];
long long p[5] = {31, 37, 41, 43, 47};
int a, b, n;
int main() {
  cin >> s >> t;
  n = t.size();
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '0')
      a++;
    else
      b++;
  }
  srand((unsigned)time(NULL));
  int k = rand() % 5;
  long long pr = p[k];
  long long cur = pr;
  for (int i = 0; i < n; i++) {
    f[i] = cur;
    hashh[i] = f[i] * (t[i] - 'a' + 1) % 1000000007;
    if (i) hashh[i] = (hashh[i] + hashh[i - 1]) % 1000000007;
    cur = cur * pr % 1000000007;
  }
  int ans = 0;
  for (int A = 1; A <= n / a; A++) {
    int zan = n - A * a;
    if (zan <= 0 || zan % b != 0) continue;
    long long B = zan / b;
    long long valA = -1, valB = -1;
    int nxt = 0;
    bool fail = 0;
    for (int j = 0; j < s.size(); j++) {
      if (s[j] == '0') {
        long long hs = hashh[nxt + A - 1];
        if (nxt) hs -= hashh[nxt - 1];
        hs = (hs % 1000000007 * f[n - 1 - nxt] % 1000000007 + 1000000007) %
             1000000007;
        nxt += A;
        if (valA == -1)
          valA = hs;
        else if (valA != hs)
          fail = 1;
      } else {
        long long hs = hashh[nxt + B - 1];
        if (nxt) hs -= hashh[nxt - 1];
        hs = (hs % 1000000007 * f[n - 1 - nxt] % 1000000007 + 1000000007) %
             1000000007;
        nxt += B;
        if (valB == -1)
          valB = hs;
        else if (valB != hs)
          fail = 1;
      }
    }
    assert(nxt == n);
    if (A == B && valA == valB) continue;
    if (!fail) ans++;
  }
  cout << ans << endl;
}

### 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;
string s, t;
long long f[1000005], hashh[1000005];
long long p[5] = {31, 37, 41, 43, 47};
int a, b, n;
int main() {
  cin >> s >> t;
  n = t.size();
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '0')
      a++;
    else
      b++;
  }
  srand((unsigned)time(NULL));
  int k = rand() % 5;
  long long pr = p[k];
  long long cur = pr;
  for (int i = 0; i < n; i++) {
    f[i] = cur;
    hashh[i] = f[i] * (t[i] - 'a' + 1) % 1000000007;
    if (i) hashh[i] = (hashh[i] + hashh[i - 1]) % 1000000007;
    cur = cur * pr % 1000000007;
  }
  int ans = 0;
  for (int A = 1; A <= n / a; A++) {
    int zan = n - A * a;
    if (zan <= 0 || zan % b != 0) continue;
    long long B = zan / b;
    long long valA = -1, valB = -1;
    int nxt = 0;
    bool fail = 0;
    for (int j = 0; j < s.size(); j++) {
      if (s[j] == '0') {
        long long hs = hashh[nxt + A - 1];
        if (nxt) hs -= hashh[nxt - 1];
        hs = (hs % 1000000007 * f[n - 1 - nxt] % 1000000007 + 1000000007) %
             1000000007;
        nxt += A;
        if (valA == -1)
          valA = hs;
        else if (valA != hs)
          fail = 1;
      } else {
        long long hs = hashh[nxt + B - 1];
        if (nxt) hs -= hashh[nxt - 1];
        hs = (hs % 1000000007 * f[n - 1 - nxt] % 1000000007 + 1000000007) %
             1000000007;
        nxt += B;
        if (valB == -1)
          valB = hs;
        else if (valB != hs)
          fail = 1;
      }
    }
    assert(nxt == n);
    if (A == B && valA == valB) continue;
    if (!fail) ans++;
  }
  cout << ans << endl;
}

```