#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const double EPS = 1e-8;
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
const long long INF = 1e18;
long long power(long long x, long long y) {
  long long t = 1;
  while (y > 0) {
    if (y % 2)
      y -= 1, t = t * x % mod;
    else
      y /= 2, x = x * x % mod;
  }
  return t;
}
char s[N], t[N];
long long pow26[N], thash[N], inv[N];
long long strhash(int st, int end) {
  long long sub;
  if (st)
    sub = thash[st - 1];
  else
    sub = 0;
  return ((thash[end - 1] - (sub * pow26[end - st]) % mod + mod)) % mod;
}
int main() {
  pow26[0] = 1, inv[0] = 1;
  for (int i = 1; i < N; ++i) {
    pow26[i] = (pow26[i - 1] * 27) % mod;
    inv[i] = (power(27, mod - 2) * inv[i - 1]) % mod;
  }
  int firstzero = -1, firstone = -1, numzeroes = 0, numones = 0, n, m,
      pairs = 0;
  long long ans = 0;
  scanf("%s%s", s, t);
  n = strlen(s), m = strlen(t);
  for (int i = 0; i < m; i++) {
    thash[i] = (27 * ans + t[i] - 'a' + 1) % mod;
    ans = thash[i];
  }
  for (int i = 0; i < n; ++i) {
    if (s[i] == '0') {
      if (firstzero == -1) firstzero = i;
      numzeroes++;
    } else {
      if (firstone == -1) firstone = i;
      numones++;
    }
  }
  for (int i = 1; i * numzeroes < m; i++) {
    if ((m - i * numzeroes) % numones == 0) {
      int j = (m - i * numzeroes) / numones;
      long long zhash = strhash(j * firstzero, j * firstzero + i);
      long long ohash = strhash(i * firstone, i * firstone + j);
      if (zhash == ohash) continue;
      long long rethash = 0;
      for (int pos = 0; pos < n; ++pos) {
        if (s[pos] == '0') {
          rethash = (pow26[i] * rethash + zhash);
        } else {
          rethash = (pow26[j] * rethash + ohash);
        }
        if (rethash > mod) rethash %= mod;
      }
      if (rethash == thash[m - 1]) pairs++;
    }
  }
  printf("%d\n", pairs);
  return 0;
}

### Prompt

Please provide a cpp coded solution 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>
#pragma GCC optimize("O3")
using namespace std;
const double EPS = 1e-8;
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
const long long INF = 1e18;
long long power(long long x, long long y) {
  long long t = 1;
  while (y > 0) {
    if (y % 2)
      y -= 1, t = t * x % mod;
    else
      y /= 2, x = x * x % mod;
  }
  return t;
}
char s[N], t[N];
long long pow26[N], thash[N], inv[N];
long long strhash(int st, int end) {
  long long sub;
  if (st)
    sub = thash[st - 1];
  else
    sub = 0;
  return ((thash[end - 1] - (sub * pow26[end - st]) % mod + mod)) % mod;
}
int main() {
  pow26[0] = 1, inv[0] = 1;
  for (int i = 1; i < N; ++i) {
    pow26[i] = (pow26[i - 1] * 27) % mod;
    inv[i] = (power(27, mod - 2) * inv[i - 1]) % mod;
  }
  int firstzero = -1, firstone = -1, numzeroes = 0, numones = 0, n, m,
      pairs = 0;
  long long ans = 0;
  scanf("%s%s", s, t);
  n = strlen(s), m = strlen(t);
  for (int i = 0; i < m; i++) {
    thash[i] = (27 * ans + t[i] - 'a' + 1) % mod;
    ans = thash[i];
  }
  for (int i = 0; i < n; ++i) {
    if (s[i] == '0') {
      if (firstzero == -1) firstzero = i;
      numzeroes++;
    } else {
      if (firstone == -1) firstone = i;
      numones++;
    }
  }
  for (int i = 1; i * numzeroes < m; i++) {
    if ((m - i * numzeroes) % numones == 0) {
      int j = (m - i * numzeroes) / numones;
      long long zhash = strhash(j * firstzero, j * firstzero + i);
      long long ohash = strhash(i * firstone, i * firstone + j);
      if (zhash == ohash) continue;
      long long rethash = 0;
      for (int pos = 0; pos < n; ++pos) {
        if (s[pos] == '0') {
          rethash = (pow26[i] * rethash + zhash);
        } else {
          rethash = (pow26[j] * rethash + ohash);
        }
        if (rethash > mod) rethash %= mod;
      }
      if (rethash == thash[m - 1]) pairs++;
    }
  }
  printf("%d\n", pairs);
  return 0;
}

```