#include <bits/stdc++.h>
using namespace std;
const int MM = 1e6 + 5, MOD = 1e9 + 7, BASE = 131;
long long N, M, pw[MM], c, z, p, ans;
vector<long long> h;
string second, t;
long long hs(int l, int r) {
  return (h[r] - (h[l - 1] * pw[r - l + 1] + MOD) % MOD + MOD) % MOD;
}
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  cin >> second >> t;
  N = second.size();
  M = t.size();
  second = "%" + second;
  t = "%" + t;
  for (int i = 1; i <= N; ++i)
    if (second[i] == '0') z++;
  if (z >= second.size() / 2)
    p = 1;
  else
    z = N - z;
  h.resize(M + 1);
  pw[0] = 1;
  for (int i = 1; i <= M; ++i) {
    pw[i] = (pw[i - 1] * BASE) % MOD;
    h[i] = (h[i - 1] * BASE + (t[i] - 'a' + 1)) % MOD;
  }
  for (int i = 1; i <= M / z; ++i) {
    long long sz1 = i, sz0 = (M - sz1 * z) / (N - z);
    if ((M - sz1 * z) % (N - z) != 0 || i == M || sz0 == 0) continue;
    set<long long> hst;
    for (int j = 1, lft = 1; j <= N; ++j) {
      if ((((second[j] - '0') + p) & 1) == 1) {
        hst.insert(hs(lft, lft + sz1 - 1));
        lft += sz1;
      } else {
        hst.insert(hs(lft, lft + sz0 - 1));
        lft += sz0;
      }
    }
    if (hst.size() == 2) ++ans;
  }
  cout << ans << "\n";
}

### 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>
using namespace std;
const int MM = 1e6 + 5, MOD = 1e9 + 7, BASE = 131;
long long N, M, pw[MM], c, z, p, ans;
vector<long long> h;
string second, t;
long long hs(int l, int r) {
  return (h[r] - (h[l - 1] * pw[r - l + 1] + MOD) % MOD + MOD) % MOD;
}
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  cin >> second >> t;
  N = second.size();
  M = t.size();
  second = "%" + second;
  t = "%" + t;
  for (int i = 1; i <= N; ++i)
    if (second[i] == '0') z++;
  if (z >= second.size() / 2)
    p = 1;
  else
    z = N - z;
  h.resize(M + 1);
  pw[0] = 1;
  for (int i = 1; i <= M; ++i) {
    pw[i] = (pw[i - 1] * BASE) % MOD;
    h[i] = (h[i - 1] * BASE + (t[i] - 'a' + 1)) % MOD;
  }
  for (int i = 1; i <= M / z; ++i) {
    long long sz1 = i, sz0 = (M - sz1 * z) / (N - z);
    if ((M - sz1 * z) % (N - z) != 0 || i == M || sz0 == 0) continue;
    set<long long> hst;
    for (int j = 1, lft = 1; j <= N; ++j) {
      if ((((second[j] - '0') + p) & 1) == 1) {
        hst.insert(hs(lft, lft + sz1 - 1));
        lft += sz1;
      } else {
        hst.insert(hs(lft, lft + sz0 - 1));
        lft += sz0;
      }
    }
    if (hst.size() == 2) ++ans;
  }
  cout << ans << "\n";
}

```