#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
const int maxm = 2000026;
const long long p1 = 307;
const long long p2 = 100000223ll;
char x[maxn], y[maxm];
int n, m;
int c0 = 0, c1 = 0;
long long hsh[maxm];
long long pwr[maxm];
bool eql(int l1, int r1, int l2, int r2) {
  return (hsh[r1] - (hsh[l1 - 1] * pwr[r1 - l1 + 1]) % p2 + p2) % p2 ==
         (hsh[r2] - (hsh[l2 - 1] * pwr[r2 - l2 + 1]) % p2 + p2) % p2;
}
int main() {
  cin >> x >> y;
  n = strlen(x);
  m = strlen(y);
  hsh[1] = y[0];
  pwr[0] = 1;
  for (int i = 1; i <= m + 2; i++) pwr[i] = (pwr[i - 1] * p1) % p2;
  for (int i = 1; i < m; i++) {
    hsh[i + 1] = (hsh[i] * p1 + y[i]) % p2;
  }
  if (x[0] == '1') {
    for (int i = 0; i < n; i++) {
      if (x[i] == '0')
        x[i] = '1';
      else
        x[i] = '0';
    }
  }
  for (int i = 0; i < n; i++) {
    if (x[i] == '0')
      c0++;
    else
      c1++;
  }
  long long ans = 0;
  for (int i = 1; i <= m; i++) {
    int ch = m - i * c0;
    if (ch <= 0) continue;
    if (ch % c1) continue;
    int l1 = ch / c1;
    int l0 = 1;
    bool ok = 1;
    int c1l = -1;
    for (int j = 0; j < n; j++) {
      if (x[j] == '0') {
        if (l0 + i - 1 > m) {
          ok = 0;
          break;
        }
        if (!eql(1, 1 + i - 1, l0, l0 + i - 1)) ok = 0;
        l0 = l0 + i;
      } else if (x[j] == '1') {
        if (l0 + l1 - 1 > m) {
          ok = 0;
          break;
        }
        if (c1l == -1) {
          c1l = l0;
          l0 = l0 + l1;
        } else {
          if (!eql(c1l, c1l + l1 - 1, l0, l0 + l1 - 1)) ok = 0;
          l0 = l0 + l1;
        }
      }
    }
    if (i == l1 && eql(1, 1 + i - 1, c1l, c1l + l1 - 1)) ok = 0;
    if (ok) ans++;
  }
  cout << ans;
  return 0;
}

### Prompt

Create a solution in Cpp for 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 int maxn = 100005;
const int maxm = 2000026;
const long long p1 = 307;
const long long p2 = 100000223ll;
char x[maxn], y[maxm];
int n, m;
int c0 = 0, c1 = 0;
long long hsh[maxm];
long long pwr[maxm];
bool eql(int l1, int r1, int l2, int r2) {
  return (hsh[r1] - (hsh[l1 - 1] * pwr[r1 - l1 + 1]) % p2 + p2) % p2 ==
         (hsh[r2] - (hsh[l2 - 1] * pwr[r2 - l2 + 1]) % p2 + p2) % p2;
}
int main() {
  cin >> x >> y;
  n = strlen(x);
  m = strlen(y);
  hsh[1] = y[0];
  pwr[0] = 1;
  for (int i = 1; i <= m + 2; i++) pwr[i] = (pwr[i - 1] * p1) % p2;
  for (int i = 1; i < m; i++) {
    hsh[i + 1] = (hsh[i] * p1 + y[i]) % p2;
  }
  if (x[0] == '1') {
    for (int i = 0; i < n; i++) {
      if (x[i] == '0')
        x[i] = '1';
      else
        x[i] = '0';
    }
  }
  for (int i = 0; i < n; i++) {
    if (x[i] == '0')
      c0++;
    else
      c1++;
  }
  long long ans = 0;
  for (int i = 1; i <= m; i++) {
    int ch = m - i * c0;
    if (ch <= 0) continue;
    if (ch % c1) continue;
    int l1 = ch / c1;
    int l0 = 1;
    bool ok = 1;
    int c1l = -1;
    for (int j = 0; j < n; j++) {
      if (x[j] == '0') {
        if (l0 + i - 1 > m) {
          ok = 0;
          break;
        }
        if (!eql(1, 1 + i - 1, l0, l0 + i - 1)) ok = 0;
        l0 = l0 + i;
      } else if (x[j] == '1') {
        if (l0 + l1 - 1 > m) {
          ok = 0;
          break;
        }
        if (c1l == -1) {
          c1l = l0;
          l0 = l0 + l1;
        } else {
          if (!eql(c1l, c1l + l1 - 1, l0, l0 + l1 - 1)) ok = 0;
          l0 = l0 + l1;
        }
      }
    }
    if (i == l1 && eql(1, 1 + i - 1, c1l, c1l + l1 - 1)) ok = 0;
    if (ok) ans++;
  }
  cout << ans;
  return 0;
}

```