#include <bits/stdc++.h>
using namespace std;
char s[110000], t[1100000];
int lens, lent;
unsigned long long base = 26;
unsigned long long p[1100000], ha[1100000];
unsigned long long val1[110000], val2[110000];
bool check(int l1, int l2) {
  int i, j, x = 0;
  int c1 = 0, c2 = 0;
  for (i = 0; i < lens; ++i) {
    if (s[i] == '0') {
      val1[++c1] = ha[x + l1] - ha[x] * p[l1];
      x += l1;
    } else {
      val2[++c2] = ha[x + l2] - ha[x] * p[l2];
      x += l2;
    }
  }
  for (i = c1; i > 1; --i)
    if (val1[i] != val1[i - 1]) return 0;
  for (i = c2; i > 1; --i)
    if (val2[i] != val2[i - 1]) return 0;
  if (l1 == l2 && val1[1] == val2[1]) return 0;
  return 1;
}
int main() {
  int i, j, cnt1, cnt0;
  int ans = 0;
  scanf("%s", s);
  scanf("%s", t);
  lens = strlen(s);
  lent = strlen(t);
  cnt0 = cnt1 = 0;
  for (i = 0; i < lens; ++i)
    if (s[i] == '0')
      ++cnt0;
    else
      ++cnt1;
  p[0] = 1;
  for (i = 1; i <= lent; ++i) p[i] = p[i - 1] * base;
  ha[1] = t[0] - 'a';
  for (i = 2; i <= lent; ++i) ha[i] = ha[i - 1] * base + t[i - 1] - 'a';
  if (cnt0 > cnt1) {
    for (i = 1; i <= lent / cnt0; ++i)
      if (cnt1 && (lent - i * cnt0) % cnt1 == 0 || !cnt1 && i * cnt0 == lent) {
        if (cnt1)
          j = (lent - i * cnt0) / cnt1;
        else
          j = 0;
        if (cnt1 && j == 0) continue;
        if (check(i, j)) ++ans;
      }
  } else
    for (i = 1; i <= lent / cnt1; ++i)
      if ((cnt0 && (lent - i * cnt1) % cnt0 == 0) ||
          (!cnt0 && i * cnt1 == lent)) {
        if (cnt0)
          j = (lent - i * cnt1) / cnt0;
        else
          j = 0;
        if (cnt0 && j == 0) continue;
        if (check(j, i)) ++ans;
      }
  printf("%d", ans);
  return 0;
}

### Prompt

Please formulate 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;
char s[110000], t[1100000];
int lens, lent;
unsigned long long base = 26;
unsigned long long p[1100000], ha[1100000];
unsigned long long val1[110000], val2[110000];
bool check(int l1, int l2) {
  int i, j, x = 0;
  int c1 = 0, c2 = 0;
  for (i = 0; i < lens; ++i) {
    if (s[i] == '0') {
      val1[++c1] = ha[x + l1] - ha[x] * p[l1];
      x += l1;
    } else {
      val2[++c2] = ha[x + l2] - ha[x] * p[l2];
      x += l2;
    }
  }
  for (i = c1; i > 1; --i)
    if (val1[i] != val1[i - 1]) return 0;
  for (i = c2; i > 1; --i)
    if (val2[i] != val2[i - 1]) return 0;
  if (l1 == l2 && val1[1] == val2[1]) return 0;
  return 1;
}
int main() {
  int i, j, cnt1, cnt0;
  int ans = 0;
  scanf("%s", s);
  scanf("%s", t);
  lens = strlen(s);
  lent = strlen(t);
  cnt0 = cnt1 = 0;
  for (i = 0; i < lens; ++i)
    if (s[i] == '0')
      ++cnt0;
    else
      ++cnt1;
  p[0] = 1;
  for (i = 1; i <= lent; ++i) p[i] = p[i - 1] * base;
  ha[1] = t[0] - 'a';
  for (i = 2; i <= lent; ++i) ha[i] = ha[i - 1] * base + t[i - 1] - 'a';
  if (cnt0 > cnt1) {
    for (i = 1; i <= lent / cnt0; ++i)
      if (cnt1 && (lent - i * cnt0) % cnt1 == 0 || !cnt1 && i * cnt0 == lent) {
        if (cnt1)
          j = (lent - i * cnt0) / cnt1;
        else
          j = 0;
        if (cnt1 && j == 0) continue;
        if (check(i, j)) ++ans;
      }
  } else
    for (i = 1; i <= lent / cnt1; ++i)
      if ((cnt0 && (lent - i * cnt1) % cnt0 == 0) ||
          (!cnt0 && i * cnt1 == lent)) {
        if (cnt0)
          j = (lent - i * cnt1) / cnt0;
        else
          j = 0;
        if (cnt0 && j == 0) continue;
        if (check(j, i)) ++ans;
      }
  printf("%d", ans);
  return 0;
}

```