#include <bits/stdc++.h>
char ch;
bool fs;
void re(int& x) {
  while (ch = getchar(), ch < 33)
    ;
  if (ch == '-')
    fs = 1, x = 0;
  else
    fs = 0, x = ch - 48;
  while (ch = getchar(), ch > 33) x = x * 10 + ch - 48;
  if (fs) x = -x;
}
using namespace std;
char s[2000001], t[2000001];
int hash1[2000001], hash2[2000001], mi1[2000001], mi2[2000001];
const int mo1 = 1000000007, mo2 = 1000000009;
int n, m, ct0, ct1, ans;
bool check(int a, int b, int c, int d) {
  return !((hash1[b] - hash1[d] -
            1ll * mi1[b - a + 1] * (hash1[a - 1] - hash1[c - 1])) %
           mo1) &&
         !((hash2[b] - hash2[d] -
            1ll * mi2[b - a + 1] * (hash2[a - 1] - hash2[c - 1])) %
           mo2);
}
int main() {
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (int i = 1; i <= n; ++i)
    if (s[i] == '0')
      ++ct0;
    else
      ++ct1;
  mi1[0] = mi2[0] = 1;
  for (int i = 1; i <= m; ++i) {
    mi1[i] = 1ll * mi1[i - 1] * 27 % mo1;
    mi2[i] = 1ll * mi2[i - 1] * 27 % mo2;
    hash1[i] = (27ll * hash1[i - 1] + t[i] - 'a' + 1) % mo1;
    hash2[i] = (27ll * hash2[i - 1] + t[i] - 'a' + 1) % mo2;
  }
  for (int i = 1; i * ct0 < m; ++i) {
    int l0 = i, l1 = (m - l0 * ct0) / ct1;
    if ((m - l0 * ct0) % ct1) continue;
    int la0 = 0, la1 = 0, now = 1;
    bool ok = 1;
    for (int j = 1; j <= n; ++j)
      if (s[j] == '0') {
        if (la0 && !check(now, now + l0 - 1, la0, la0 + l0 - 1)) {
          ok = 0;
          break;
        }
        la0 = now;
        now += l0;
      } else {
        if (la1 && !check(now, now + l1 - 1, la1, la1 + l1 - 1)) {
          ok = 0;
          break;
        }
        la1 = now;
        now += l1;
      }
    if (!ok) continue;
    if (!check(la0, la0 + l0 - 1, la1, la1 + l1 - 1)) ++ans;
  }
  printf("%d\n", ans);
}

### Prompt

Develop a solution in Cpp 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>
char ch;
bool fs;
void re(int& x) {
  while (ch = getchar(), ch < 33)
    ;
  if (ch == '-')
    fs = 1, x = 0;
  else
    fs = 0, x = ch - 48;
  while (ch = getchar(), ch > 33) x = x * 10 + ch - 48;
  if (fs) x = -x;
}
using namespace std;
char s[2000001], t[2000001];
int hash1[2000001], hash2[2000001], mi1[2000001], mi2[2000001];
const int mo1 = 1000000007, mo2 = 1000000009;
int n, m, ct0, ct1, ans;
bool check(int a, int b, int c, int d) {
  return !((hash1[b] - hash1[d] -
            1ll * mi1[b - a + 1] * (hash1[a - 1] - hash1[c - 1])) %
           mo1) &&
         !((hash2[b] - hash2[d] -
            1ll * mi2[b - a + 1] * (hash2[a - 1] - hash2[c - 1])) %
           mo2);
}
int main() {
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (int i = 1; i <= n; ++i)
    if (s[i] == '0')
      ++ct0;
    else
      ++ct1;
  mi1[0] = mi2[0] = 1;
  for (int i = 1; i <= m; ++i) {
    mi1[i] = 1ll * mi1[i - 1] * 27 % mo1;
    mi2[i] = 1ll * mi2[i - 1] * 27 % mo2;
    hash1[i] = (27ll * hash1[i - 1] + t[i] - 'a' + 1) % mo1;
    hash2[i] = (27ll * hash2[i - 1] + t[i] - 'a' + 1) % mo2;
  }
  for (int i = 1; i * ct0 < m; ++i) {
    int l0 = i, l1 = (m - l0 * ct0) / ct1;
    if ((m - l0 * ct0) % ct1) continue;
    int la0 = 0, la1 = 0, now = 1;
    bool ok = 1;
    for (int j = 1; j <= n; ++j)
      if (s[j] == '0') {
        if (la0 && !check(now, now + l0 - 1, la0, la0 + l0 - 1)) {
          ok = 0;
          break;
        }
        la0 = now;
        now += l0;
      } else {
        if (la1 && !check(now, now + l1 - 1, la1, la1 + l1 - 1)) {
          ok = 0;
          break;
        }
        la1 = now;
        now += l1;
      }
    if (!ok) continue;
    if (!check(la0, la0 + l0 - 1, la1, la1 + l1 - 1)) ++ans;
  }
  printf("%d\n", ans);
}

```