#include <bits/stdc++.h>
using namespace std;
string S, T;
long long ans, HS[10000002], lt[10000002];
long long geths(long long x, long long y) {
  return (HS[y + 1] - ((HS[x] * lt[y - x + 1]) % 1000000007) + 1000000007) %
         1000000007;
}
void input() { cin >> S >> T; }
void output() { cout << ans; }
signed main() {
  if (fopen("GHour"
            ".inp",
            "r")) {
    freopen(
        "GHour"
        ".inp",
        "r", stdin);
    freopen(
        "GHour"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  input();
  lt[0] = 1;
  for (long long i = 1; i <= T.size(); i++) {
    lt[i] = (lt[i - 1] * 29) % 1000000007;
  }
  HS[0] = 0;
  for (long long i = 0; i < T.size(); i++) {
    HS[i + 1] =
        (((HS[i] * 29) % 1000000007) + (long long)T[i] - 96) % 1000000007;
  }
  long long num0 = 0;
  long long num1 = 0;
  for (long long i = 0; i < S.size(); i++) {
    if (S[i] == '0') {
      num0++;
    } else if (S[i] == '1') {
      num1++;
    }
  }
  for (long long i = 1; i <= T.size(); i++) {
    if ((((T.size() - (num0 * i)) % num1) || ((num0 * i) > T.size()))) continue;
    long long x = i;
    long long y = (T.size() - (num0 * i)) / num1;
    if (y == 0) continue;
    long long runS = 0;
    long long runT = 0;
    long long check0 = 0;
    long long check1 = 0;
    long long check = 0;
    long long str0 = 0;
    long long str1 = 0;
    while (runS < S.size() && runT < T.size()) {
      if (S[runS] == '0') {
        if (check0 == 0) {
          str0 = geths(runT, runT + x - 1);
          check0 = 1;
        } else {
          if (geths(runT, runT + x - 1) != str0) {
            check = 1;
            break;
          }
        }
        runT += x;
      } else if (S[runS] == '1') {
        if (check1 == 0) {
          str1 = geths(runT, runT + y - 1);
          check1 = 1;
        } else {
          if (geths(runT, runT + y - 1) != str1) {
            check = 1;
            break;
          }
        }
        runT += y;
      }
      runS++;
    }
    if (check == 1 || str0 == str1) continue;
    ans++;
  }
  output();
  return 0;
}

### 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 ans, HS[10000002], lt[10000002];
long long geths(long long x, long long y) {
  return (HS[y + 1] - ((HS[x] * lt[y - x + 1]) % 1000000007) + 1000000007) %
         1000000007;
}
void input() { cin >> S >> T; }
void output() { cout << ans; }
signed main() {
  if (fopen("GHour"
            ".inp",
            "r")) {
    freopen(
        "GHour"
        ".inp",
        "r", stdin);
    freopen(
        "GHour"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  input();
  lt[0] = 1;
  for (long long i = 1; i <= T.size(); i++) {
    lt[i] = (lt[i - 1] * 29) % 1000000007;
  }
  HS[0] = 0;
  for (long long i = 0; i < T.size(); i++) {
    HS[i + 1] =
        (((HS[i] * 29) % 1000000007) + (long long)T[i] - 96) % 1000000007;
  }
  long long num0 = 0;
  long long num1 = 0;
  for (long long i = 0; i < S.size(); i++) {
    if (S[i] == '0') {
      num0++;
    } else if (S[i] == '1') {
      num1++;
    }
  }
  for (long long i = 1; i <= T.size(); i++) {
    if ((((T.size() - (num0 * i)) % num1) || ((num0 * i) > T.size()))) continue;
    long long x = i;
    long long y = (T.size() - (num0 * i)) / num1;
    if (y == 0) continue;
    long long runS = 0;
    long long runT = 0;
    long long check0 = 0;
    long long check1 = 0;
    long long check = 0;
    long long str0 = 0;
    long long str1 = 0;
    while (runS < S.size() && runT < T.size()) {
      if (S[runS] == '0') {
        if (check0 == 0) {
          str0 = geths(runT, runT + x - 1);
          check0 = 1;
        } else {
          if (geths(runT, runT + x - 1) != str0) {
            check = 1;
            break;
          }
        }
        runT += x;
      } else if (S[runS] == '1') {
        if (check1 == 0) {
          str1 = geths(runT, runT + y - 1);
          check1 = 1;
        } else {
          if (geths(runT, runT + y - 1) != str1) {
            check = 1;
            break;
          }
        }
        runT += y;
      }
      runS++;
    }
    if (check == 1 || str0 == str1) continue;
    ans++;
  }
  output();
  return 0;
}

```