#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
using namespace std;
const int N = 1e6 + 5;
const int MOD = 1e9 + 7;
const int base = 31;
char s[N], t[N];
long long h[N], p[N];
int n, m, c0 = 0, c1 = 0;
long long get_hash(int l, int r) {
  return ((h[r] - h[l - 1] * p[r - l + 1]) % MOD + MOD) % MOD;
}
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')
      c0++;
    else
      c1++;
  p[0] = 1;
  for (int i = 1; i <= m; ++i)
    h[i] = (h[i - 1] * base + t[i] - 'a' + 1) % MOD,
    p[i] = (p[i - 1] * base) % MOD;
  int ans = 0;
  if (c0 == 0) {
    if (m % c1 == 0) {
      unsigned long long h = 0;
      bool f = true;
      for (int i = c1; i <= m; i += c1) {
        if (h == 0)
          h = get_hash(i - c1 + 1, i);
        else {
          if (get_hash(i - c1 + 1, i) != h) {
            f = false;
            break;
          }
        }
      }
      if (f) ans++;
    }
  } else if (c1 == 0) {
    if (m % c0 == 0) {
      long long h = 0;
      bool f = true;
      for (int i = c0; i <= m; i += c0) {
        if (h == 0)
          h = get_hash(i - c0 + 1, i);
        else {
          if (get_hash(i - c0 + 1, i) != h) {
            f = false;
            break;
          }
        }
      }
      if (f) ans++;
    }
  } else {
    for (int i = 1; i * c1 < m; ++i) {
      if ((m - i * c1) % c0 == 0) {
        bool f = true;
        int l1 = i, l0 = (m - i * c1) / c0;
        long long h0 = 0, h1 = 0;
        int now = 1;
        for (int i = 1; i <= n; ++i) {
          if (s[i] == '0') {
            if (h0 == 0)
              h0 = get_hash(now, now + l0 - 1);
            else {
              if (get_hash(now, now + l0 - 1) != h0) {
                f = false;
                break;
              }
            }
            now += l0;
          } else {
            if (h1 == 0)
              h1 = get_hash(now, now + l1 - 1);
            else {
              if (get_hash(now, now + l1 - 1) != h1) {
                f = false;
                break;
              }
            }
            now += l1;
          }
        }
        if (f && h0 != h1) ans++;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
using namespace std;
const int N = 1e6 + 5;
const int MOD = 1e9 + 7;
const int base = 31;
char s[N], t[N];
long long h[N], p[N];
int n, m, c0 = 0, c1 = 0;
long long get_hash(int l, int r) {
  return ((h[r] - h[l - 1] * p[r - l + 1]) % MOD + MOD) % MOD;
}
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')
      c0++;
    else
      c1++;
  p[0] = 1;
  for (int i = 1; i <= m; ++i)
    h[i] = (h[i - 1] * base + t[i] - 'a' + 1) % MOD,
    p[i] = (p[i - 1] * base) % MOD;
  int ans = 0;
  if (c0 == 0) {
    if (m % c1 == 0) {
      unsigned long long h = 0;
      bool f = true;
      for (int i = c1; i <= m; i += c1) {
        if (h == 0)
          h = get_hash(i - c1 + 1, i);
        else {
          if (get_hash(i - c1 + 1, i) != h) {
            f = false;
            break;
          }
        }
      }
      if (f) ans++;
    }
  } else if (c1 == 0) {
    if (m % c0 == 0) {
      long long h = 0;
      bool f = true;
      for (int i = c0; i <= m; i += c0) {
        if (h == 0)
          h = get_hash(i - c0 + 1, i);
        else {
          if (get_hash(i - c0 + 1, i) != h) {
            f = false;
            break;
          }
        }
      }
      if (f) ans++;
    }
  } else {
    for (int i = 1; i * c1 < m; ++i) {
      if ((m - i * c1) % c0 == 0) {
        bool f = true;
        int l1 = i, l0 = (m - i * c1) / c0;
        long long h0 = 0, h1 = 0;
        int now = 1;
        for (int i = 1; i <= n; ++i) {
          if (s[i] == '0') {
            if (h0 == 0)
              h0 = get_hash(now, now + l0 - 1);
            else {
              if (get_hash(now, now + l0 - 1) != h0) {
                f = false;
                break;
              }
            }
            now += l0;
          } else {
            if (h1 == 0)
              h1 = get_hash(now, now + l1 - 1);
            else {
              if (get_hash(now, now + l1 - 1) != h1) {
                f = false;
                break;
              }
            }
            now += l1;
          }
        }
        if (f && h0 != h1) ans++;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

```