#include <bits/stdc++.h>
char s[1000010], t[1000010];
int n, m;
const int moha = 19260817;
const int mod = 1e9 + 7;
int pre1[1000010], pre2[1000010], pw1[1000010], pw2[1000010];
inline bool check(int l0, int l1) {
  int l = 1, r = 0;
  int hash0_1 = -1, hash0_2 = -1, hash1_1 = -1, hash1_2 = -1;
  for (int i = 1; i <= n; ++i) {
    l = r + 1;
    r = l + (s[i] == '0' ? l0 : l1) - 1;
    int key1 =
        (pre1[r] - 1ll * pre1[l - 1] * pw1[r - l + 1] % moha + moha) % moha;
    int key2 = (pre2[r] - 1ll * pre2[l - 1] * pw2[r - l + 1] % mod + mod) % mod;
    if (s[i] == '0') {
      if (hash0_1 == -1)
        hash0_1 = key1, hash0_2 = key2;
      else if (hash0_1 != key1 || hash0_2 != key2)
        return 0;
    } else {
      if (hash1_1 == -1)
        hash1_1 = key1, hash1_2 = key2;
      else if (hash1_1 != key1 || hash1_2 != key2)
        return 0;
    }
  }
  return !(hash0_1 == hash1_1 && hash0_2 == hash1_2);
}
int main() {
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1), m = strlen(t + 1);
  pw1[0] = pw2[0] = 1;
  for (int i = 1; i <= m; ++i) {
    pw1[i] = 29ll * pw1[i - 1] % moha;
    pw2[i] = 29ll * pw2[i - 1] % mod;
    pre1[i] = (1ll * pre1[i - 1] * 29 + t[i] - 'a' + 1) % moha;
    pre2[i] = (1ll * pre2[i - 1] * 29 + t[i] - 'a' + 1) % mod;
  }
  int c0 = 0, c1 = 0;
  for (int i = 1; i <= n; ++i) c0 += s[i] == '0', c1 += s[i] == '1';
  int ans = 0;
  for (int i = 1; i <= m && i * c0 < m; ++i)
    if ((m - i * c0) % c1 == 0) {
      int j = (m - i * c0) / c1;
      if (check(i, j)) ++ans;
    }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate 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>
char s[1000010], t[1000010];
int n, m;
const int moha = 19260817;
const int mod = 1e9 + 7;
int pre1[1000010], pre2[1000010], pw1[1000010], pw2[1000010];
inline bool check(int l0, int l1) {
  int l = 1, r = 0;
  int hash0_1 = -1, hash0_2 = -1, hash1_1 = -1, hash1_2 = -1;
  for (int i = 1; i <= n; ++i) {
    l = r + 1;
    r = l + (s[i] == '0' ? l0 : l1) - 1;
    int key1 =
        (pre1[r] - 1ll * pre1[l - 1] * pw1[r - l + 1] % moha + moha) % moha;
    int key2 = (pre2[r] - 1ll * pre2[l - 1] * pw2[r - l + 1] % mod + mod) % mod;
    if (s[i] == '0') {
      if (hash0_1 == -1)
        hash0_1 = key1, hash0_2 = key2;
      else if (hash0_1 != key1 || hash0_2 != key2)
        return 0;
    } else {
      if (hash1_1 == -1)
        hash1_1 = key1, hash1_2 = key2;
      else if (hash1_1 != key1 || hash1_2 != key2)
        return 0;
    }
  }
  return !(hash0_1 == hash1_1 && hash0_2 == hash1_2);
}
int main() {
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1), m = strlen(t + 1);
  pw1[0] = pw2[0] = 1;
  for (int i = 1; i <= m; ++i) {
    pw1[i] = 29ll * pw1[i - 1] % moha;
    pw2[i] = 29ll * pw2[i - 1] % mod;
    pre1[i] = (1ll * pre1[i - 1] * 29 + t[i] - 'a' + 1) % moha;
    pre2[i] = (1ll * pre2[i - 1] * 29 + t[i] - 'a' + 1) % mod;
  }
  int c0 = 0, c1 = 0;
  for (int i = 1; i <= n; ++i) c0 += s[i] == '0', c1 += s[i] == '1';
  int ans = 0;
  for (int i = 1; i <= m && i * c0 < m; ++i)
    if ((m - i * c0) % c1 == 0) {
      int j = (m - i * c0) / c1;
      if (check(i, j)) ++ans;
    }
  printf("%d\n", ans);
  return 0;
}

```