#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10;
const int mod = 1e9 + 7;
const int P = 257;
int H[MAXN];
char s[MAXN];
char t[MAXN];
int iP[MAXN];
int add(int a, int b) { return (a + b) % mod; }
int mul(long long a, long long b) { return (a * b) % mod; }
int range(int l, int r) {
  if (l == 0) return H[r];
  return mul(add(H[r], mod - H[l - 1]), iP[l]);
}
int n, m;
bool check(int x, int y) {
  int pos = 0;
  int hx = -1, hy = -1;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '1') {
      if (hx == -1)
        hx = range(pos, pos + x - 1);
      else if (hx != range(pos, pos + x - 1))
        return 0;
      pos += x;
    } else {
      if (hy == -1)
        hy = range(pos, pos + y - 1);
      else if (hy != range(pos, pos + y - 1))
        return 0;
      pos += y;
    }
  }
  if (x == y) return hx != hy;
  return 1;
}
int ex(int a, int b) {
  int r = 1;
  while (b > 0) {
    if (b & 1) r = mul(r, a);
    a = mul(a, a);
    b >>= 1;
  }
  return r;
}
int main() {
  scanf("%s", s);
  scanf("%s", t);
  n = strlen(s);
  m = strlen(t);
  int c1 = 0, c0 = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '0')
      c0 += 1;
    else
      c1 += 1;
  }
  int h = 0;
  int ip = ex(P, mod - 2);
  int p = 1;
  for (int i = 0; i < m; ++i) {
    h = add(h, mul(t[i], p));
    p = mul(p, P);
    H[i] = h;
  }
  iP[0] = 1;
  for (int i = 1; i < m; ++i) {
    iP[i] = mul(iP[i - 1], ip);
  }
  int ans = 0;
  for (int x = 1; x * c1 < m; ++x) {
    if ((m - c1 * x) % c0 == 0) {
      int y = (m - c1 * x) / c0;
      ans += check(x, y);
    }
  }
  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>
using namespace std;
const int MAXN = 1e6 + 10;
const int mod = 1e9 + 7;
const int P = 257;
int H[MAXN];
char s[MAXN];
char t[MAXN];
int iP[MAXN];
int add(int a, int b) { return (a + b) % mod; }
int mul(long long a, long long b) { return (a * b) % mod; }
int range(int l, int r) {
  if (l == 0) return H[r];
  return mul(add(H[r], mod - H[l - 1]), iP[l]);
}
int n, m;
bool check(int x, int y) {
  int pos = 0;
  int hx = -1, hy = -1;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '1') {
      if (hx == -1)
        hx = range(pos, pos + x - 1);
      else if (hx != range(pos, pos + x - 1))
        return 0;
      pos += x;
    } else {
      if (hy == -1)
        hy = range(pos, pos + y - 1);
      else if (hy != range(pos, pos + y - 1))
        return 0;
      pos += y;
    }
  }
  if (x == y) return hx != hy;
  return 1;
}
int ex(int a, int b) {
  int r = 1;
  while (b > 0) {
    if (b & 1) r = mul(r, a);
    a = mul(a, a);
    b >>= 1;
  }
  return r;
}
int main() {
  scanf("%s", s);
  scanf("%s", t);
  n = strlen(s);
  m = strlen(t);
  int c1 = 0, c0 = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '0')
      c0 += 1;
    else
      c1 += 1;
  }
  int h = 0;
  int ip = ex(P, mod - 2);
  int p = 1;
  for (int i = 0; i < m; ++i) {
    h = add(h, mul(t[i], p));
    p = mul(p, P);
    H[i] = h;
  }
  iP[0] = 1;
  for (int i = 1; i < m; ++i) {
    iP[i] = mul(iP[i - 1], ip);
  }
  int ans = 0;
  for (int x = 1; x * c1 < m; ++x) {
    if ((m - c1 * x) % c0 == 0) {
      int y = (m - c1 * x) / c0;
      ans += check(x, y);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```