#include <bits/stdc++.h>
const int mod = 1000000007;
const int seed = 131;
char s[100005], t[1000005];
int hshv[1000005];
int powm(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1LL * ret * a % mod;
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return ret;
}
void gethash(char* s) {
  int len = strlen(s), now;
  now = 0;
  for (int i = 0; i < len; i++) {
    now = (1LL * now * seed + s[i]) % mod;
    hshv[i] = now;
  }
}
int get(int s, int t) {
  int a, b;
  if (s == 0)
    a = 0;
  else
    a = 1LL * hshv[s - 1] * powm(seed, t - s + 1) % mod;
  b = hshv[t];
  return (b - a + mod) % mod;
}
int main() {
  int cnt0, cnt1, lens, lent, len0, len1, hsh0, hsh1, p, ans;
  bool f;
  scanf("%s", s);
  scanf("%s", t);
  lens = strlen(s);
  lent = strlen(t);
  if (s[0] == '1')
    for (int i = 0; i < lens; i++) s[i] = ((s[i] - '0') ^ 1) + '0';
  cnt0 = 0;
  cnt1 = 0;
  for (int i = 0; i < lens; i++)
    if (s[i] == '0')
      cnt0++;
    else
      cnt1++;
  ans = 0;
  gethash(t);
  for (int i = 1; cnt0 * i < lent; i++) {
    if ((lent - cnt0 * i) % cnt1 != 0) continue;
    len0 = i;
    len1 = (lent - cnt0 * len0) / cnt1;
    hsh0 = -1;
    hsh1 = -1;
    p = 0;
    f = true;
    for (int j = 0; j < lens; j++) {
      if (s[j] == '0') {
        if (hsh0 == -1)
          hsh0 = get(p, p + len0 - 1);
        else if (get(p, p + len0 - 1) != hsh0) {
          f = false;
          break;
        }
        p += len0;
      } else {
        if (hsh1 == -1)
          hsh1 = get(p, p + len1 - 1);
        else if (get(p, p + len1 - 1) != hsh1) {
          f = false;
          break;
        }
        p += len1;
      }
    }
    if (f && hsh0 != hsh1) ans++;
  }
  printf("%d\n", ans);
}

### 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>
const int mod = 1000000007;
const int seed = 131;
char s[100005], t[1000005];
int hshv[1000005];
int powm(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1LL * ret * a % mod;
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return ret;
}
void gethash(char* s) {
  int len = strlen(s), now;
  now = 0;
  for (int i = 0; i < len; i++) {
    now = (1LL * now * seed + s[i]) % mod;
    hshv[i] = now;
  }
}
int get(int s, int t) {
  int a, b;
  if (s == 0)
    a = 0;
  else
    a = 1LL * hshv[s - 1] * powm(seed, t - s + 1) % mod;
  b = hshv[t];
  return (b - a + mod) % mod;
}
int main() {
  int cnt0, cnt1, lens, lent, len0, len1, hsh0, hsh1, p, ans;
  bool f;
  scanf("%s", s);
  scanf("%s", t);
  lens = strlen(s);
  lent = strlen(t);
  if (s[0] == '1')
    for (int i = 0; i < lens; i++) s[i] = ((s[i] - '0') ^ 1) + '0';
  cnt0 = 0;
  cnt1 = 0;
  for (int i = 0; i < lens; i++)
    if (s[i] == '0')
      cnt0++;
    else
      cnt1++;
  ans = 0;
  gethash(t);
  for (int i = 1; cnt0 * i < lent; i++) {
    if ((lent - cnt0 * i) % cnt1 != 0) continue;
    len0 = i;
    len1 = (lent - cnt0 * len0) / cnt1;
    hsh0 = -1;
    hsh1 = -1;
    p = 0;
    f = true;
    for (int j = 0; j < lens; j++) {
      if (s[j] == '0') {
        if (hsh0 == -1)
          hsh0 = get(p, p + len0 - 1);
        else if (get(p, p + len0 - 1) != hsh0) {
          f = false;
          break;
        }
        p += len0;
      } else {
        if (hsh1 == -1)
          hsh1 = get(p, p + len1 - 1);
        else if (get(p, p + len1 - 1) != hsh1) {
          f = false;
          break;
        }
        p += len1;
      }
    }
    if (f && hsh0 != hsh1) ans++;
  }
  printf("%d\n", ans);
}

```