#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007, base = 36161;
const int MAX_T = 1000007;
long long prefHash[MAX_T], Pow[MAX_T];
string s, t;
long long zeroes = 0, ones = 0, firstOne = -1, firstZero = -1, n;
long long GetHash(int a, int b);
void HashPreprocessing();
bool Check(long long a, long long b);
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s >> t;
  if (s[0] == '1')
    for (char &c : s) c = c == '0' ? '1' : '0';
  for (int i = 0; i < (int)s.size(); i++)
    if (s[i] == '0') {
      zeroes++;
      if (firstZero == -1) firstZero = i;
    } else {
      ones++;
      if (firstOne == -1) firstOne = i;
    }
  n = t.size();
  HashPreprocessing();
  long long result = 0;
  for (int a = 1; a < n; a++)
    if (n - zeroes * a > 0 && (n - zeroes * a) % ones == 0)
      result += Check(a, (n - zeroes * a) / ones);
  cout << result << "\n";
  return 0;
}
bool Check(long long a, long long b) {
  if (a * firstOne + b > n) return 0;
  long long aHash = GetHash(0, a - 1);
  long long bHash = GetHash(a * firstOne, a * firstOne + b - 1);
  if (aHash == bHash && a == b) return 0;
  long long pos = 0;
  for (char c : s) {
    if (c == '0') {
      if (aHash != GetHash(pos, pos + a - 1)) return 0;
      pos += a;
    } else {
      if (bHash != GetHash(pos, pos + b - 1)) return 0;
      pos += b;
    }
  }
  return 1;
}
void HashPreprocessing() {
  Pow[0] = 1;
  for (int i = 1; i <= n; i++) Pow[i] = (Pow[i - 1] * base) % mod;
  prefHash[0] = t[0] - 'a';
  for (int i = 1; i < n; i++)
    prefHash[i] = (prefHash[i - 1] * base + t[i] - 'a') % mod;
}
long long GetHash(int a, int b) {
  long long left = a == 0 ? 0 : (prefHash[a - 1] * Pow[b - a + 1]) % mod;
  return (prefHash[b] + mod - left) % mod;
}

### Prompt

Your task is to create 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 mod = 1000000007, base = 36161;
const int MAX_T = 1000007;
long long prefHash[MAX_T], Pow[MAX_T];
string s, t;
long long zeroes = 0, ones = 0, firstOne = -1, firstZero = -1, n;
long long GetHash(int a, int b);
void HashPreprocessing();
bool Check(long long a, long long b);
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s >> t;
  if (s[0] == '1')
    for (char &c : s) c = c == '0' ? '1' : '0';
  for (int i = 0; i < (int)s.size(); i++)
    if (s[i] == '0') {
      zeroes++;
      if (firstZero == -1) firstZero = i;
    } else {
      ones++;
      if (firstOne == -1) firstOne = i;
    }
  n = t.size();
  HashPreprocessing();
  long long result = 0;
  for (int a = 1; a < n; a++)
    if (n - zeroes * a > 0 && (n - zeroes * a) % ones == 0)
      result += Check(a, (n - zeroes * a) / ones);
  cout << result << "\n";
  return 0;
}
bool Check(long long a, long long b) {
  if (a * firstOne + b > n) return 0;
  long long aHash = GetHash(0, a - 1);
  long long bHash = GetHash(a * firstOne, a * firstOne + b - 1);
  if (aHash == bHash && a == b) return 0;
  long long pos = 0;
  for (char c : s) {
    if (c == '0') {
      if (aHash != GetHash(pos, pos + a - 1)) return 0;
      pos += a;
    } else {
      if (bHash != GetHash(pos, pos + b - 1)) return 0;
      pos += b;
    }
  }
  return 1;
}
void HashPreprocessing() {
  Pow[0] = 1;
  for (int i = 1; i <= n; i++) Pow[i] = (Pow[i - 1] * base) % mod;
  prefHash[0] = t[0] - 'a';
  for (int i = 1; i < n; i++)
    prefHash[i] = (prefHash[i - 1] * base + t[i] - 'a') % mod;
}
long long GetHash(int a, int b) {
  long long left = a == 0 ? 0 : (prefHash[a - 1] * Pow[b - a + 1]) % mod;
  return (prefHash[b] + mod - left) % mod;
}

```