#include <bits/stdc++.h>
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline void sub(long long &a, long long b) {
  a -= b;
  if (a < 0) a += 1000000007;
}
inline void add(long long &a, long long b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
template <typename T>
inline T const &MAX(T const &a, T const &b) {
  return a > b ? a : b;
}
template <typename T>
inline T const &MIN(T const &a, T const &b) {
  return a < b ? a : b;
}
inline long long qp(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000007;
    a = a * a % 1000000007, b >>= 1;
  }
  return ans;
}
inline long long qp(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % c;
    a = a * a % c, b >>= 1;
  }
  return ans;
}
using namespace std;
const unsigned long long ba = 233;
const double eps = 1e-5;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int N = 1000000 + 10, maxn = 2000000 + 10, inf = 0x3f3f3f3f;
char s[N], t[N];
long long p[N], a[N], te[2];
int num[2];
long long cal(int l, int r) {
  return (a[r] - a[l - 1] * p[r - l + 1] % 1000000007 + 1000000007) %
         1000000007;
}
int main() {
  p[0] = 1;
  for (int i = 1; i < N; i++) p[i] = p[i - 1] * 233 % 1000000007;
  scanf("%s%s", s + 1, t + 1);
  int n = strlen(s + 1), m = strlen(t + 1);
  for (int i = 1; i <= n; i++) num[s[i] - '0']++;
  for (int i = 1; i <= m; i++)
    a[i] = (a[i - 1] * 233 + t[i] - 'a' + 1) % 1000000007;
  int ans = 0, f = s[1] - '0';
  for (int l = 1; l * num[f] < m; l++) {
    int r = m - l * num[f];
    if (r % num[f ^ 1] != 0) continue;
    r /= num[f ^ 1];
    int res = 0;
    for (int i = 1; i <= n; i++) {
      if (s[i] - '0' == f)
        res++;
      else
        break;
    }
    te[f] = cal(1, l), te[f ^ 1] = cal(res * l + 1, res * l + r);
    if (te[f] == te[f ^ 1]) continue;
    bool ok = 1;
    for (int i = 1, j = 1; i <= n; i++) {
      int tel = l, ter = r;
      if (f) swap(tel, ter);
      if (s[i] == '0') {
        if (cal(j, j + tel - 1) != te[0]) {
          ok = 0;
          break;
        } else
          j += tel;
      } else {
        if (cal(j, j + ter - 1) != te[1]) {
          ok = 0;
          break;
        } else
          j += ter;
      }
    }
    ans += ok;
  }
  printf("%d\n", ans);
  return 0;
}

### 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>
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline void sub(long long &a, long long b) {
  a -= b;
  if (a < 0) a += 1000000007;
}
inline void add(long long &a, long long b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
template <typename T>
inline T const &MAX(T const &a, T const &b) {
  return a > b ? a : b;
}
template <typename T>
inline T const &MIN(T const &a, T const &b) {
  return a < b ? a : b;
}
inline long long qp(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000007;
    a = a * a % 1000000007, b >>= 1;
  }
  return ans;
}
inline long long qp(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % c;
    a = a * a % c, b >>= 1;
  }
  return ans;
}
using namespace std;
const unsigned long long ba = 233;
const double eps = 1e-5;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int N = 1000000 + 10, maxn = 2000000 + 10, inf = 0x3f3f3f3f;
char s[N], t[N];
long long p[N], a[N], te[2];
int num[2];
long long cal(int l, int r) {
  return (a[r] - a[l - 1] * p[r - l + 1] % 1000000007 + 1000000007) %
         1000000007;
}
int main() {
  p[0] = 1;
  for (int i = 1; i < N; i++) p[i] = p[i - 1] * 233 % 1000000007;
  scanf("%s%s", s + 1, t + 1);
  int n = strlen(s + 1), m = strlen(t + 1);
  for (int i = 1; i <= n; i++) num[s[i] - '0']++;
  for (int i = 1; i <= m; i++)
    a[i] = (a[i - 1] * 233 + t[i] - 'a' + 1) % 1000000007;
  int ans = 0, f = s[1] - '0';
  for (int l = 1; l * num[f] < m; l++) {
    int r = m - l * num[f];
    if (r % num[f ^ 1] != 0) continue;
    r /= num[f ^ 1];
    int res = 0;
    for (int i = 1; i <= n; i++) {
      if (s[i] - '0' == f)
        res++;
      else
        break;
    }
    te[f] = cal(1, l), te[f ^ 1] = cal(res * l + 1, res * l + r);
    if (te[f] == te[f ^ 1]) continue;
    bool ok = 1;
    for (int i = 1, j = 1; i <= n; i++) {
      int tel = l, ter = r;
      if (f) swap(tel, ter);
      if (s[i] == '0') {
        if (cal(j, j + tel - 1) != te[0]) {
          ok = 0;
          break;
        } else
          j += tel;
      } else {
        if (cal(j, j + ter - 1) != te[1]) {
          ok = 0;
          break;
        } else
          j += ter;
      }
    }
    ans += ok;
  }
  printf("%d\n", ans);
  return 0;
}

```