#include <bits/stdc++.h>
using namespace std;
struct Mod {
  const long long M;
  Mod(long long M) : M(M) {}
  long long add(long long a, long long b) { return ((a + b) % M + M) % M; }
  long long mul(long long a, long long b) { return a * b % M; }
};
struct PowMod : vector<long long>, Mod {
  PowMod(long long P, long long N, long long M)
      : vector<long long>(N, 1), Mod(M) {
    for (int i = 1; i < size(); ++i) at(i) = mul(at(i - 1), P);
  }
};
struct Hash : PowMod {
  vector<long long> f;
  Hash(const string &s, long long M, long long P = 131)
      : f(s.size() + 1), PowMod(P, s.size() + 1, M) {
    for (int i = 0; i < s.size(); ++i) f[i + 1] = (f[i] * at(1) + s[i]) % M;
  }
  long long ask(int pos, int len) {
    return add(f[pos + len], -f[pos] * at(len));
  }
};
string s, t;
int cnt[2], ans;
int main() {
  cin >> s >> t;
  for (int i = 0; i < s.size(); ++i) ++cnt[s[i] - '0'];
  Hash h(t, 1e9 + 7);
  for (int len[2] = {1}; len[0] * cnt[0] < t.size(); ++len[0]) {
    len[1] = (t.size() - cnt[0] * len[0]) / cnt[1];
    if (len[1] == 0) break;
    if (len[0] * cnt[0] + len[1] * cnt[1] == t.size()) {
      long long r[2] = {-1, -1};
      for (int i = 0, j = 0; i < s.size(); ++i) {
        if (r[s[i] - '0'] < 0) r[s[i] - '0'] = h.ask(j, len[s[i] - '0']);
        if (r[s[i] - '0'] == h.ask(j, len[s[i] - '0']))
          j += len[s[i] - '0'];
        else {
          r[0] = r[1];
          break;
        }
      }
      if (r[1] != r[0]) ++ans;
    }
  }
  cout << ans;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
struct Mod {
  const long long M;
  Mod(long long M) : M(M) {}
  long long add(long long a, long long b) { return ((a + b) % M + M) % M; }
  long long mul(long long a, long long b) { return a * b % M; }
};
struct PowMod : vector<long long>, Mod {
  PowMod(long long P, long long N, long long M)
      : vector<long long>(N, 1), Mod(M) {
    for (int i = 1; i < size(); ++i) at(i) = mul(at(i - 1), P);
  }
};
struct Hash : PowMod {
  vector<long long> f;
  Hash(const string &s, long long M, long long P = 131)
      : f(s.size() + 1), PowMod(P, s.size() + 1, M) {
    for (int i = 0; i < s.size(); ++i) f[i + 1] = (f[i] * at(1) + s[i]) % M;
  }
  long long ask(int pos, int len) {
    return add(f[pos + len], -f[pos] * at(len));
  }
};
string s, t;
int cnt[2], ans;
int main() {
  cin >> s >> t;
  for (int i = 0; i < s.size(); ++i) ++cnt[s[i] - '0'];
  Hash h(t, 1e9 + 7);
  for (int len[2] = {1}; len[0] * cnt[0] < t.size(); ++len[0]) {
    len[1] = (t.size() - cnt[0] * len[0]) / cnt[1];
    if (len[1] == 0) break;
    if (len[0] * cnt[0] + len[1] * cnt[1] == t.size()) {
      long long r[2] = {-1, -1};
      for (int i = 0, j = 0; i < s.size(); ++i) {
        if (r[s[i] - '0'] < 0) r[s[i] - '0'] = h.ask(j, len[s[i] - '0']);
        if (r[s[i] - '0'] == h.ask(j, len[s[i] - '0']))
          j += len[s[i] - '0'];
        else {
          r[0] = r[1];
          break;
        }
      }
      if (r[1] != r[0]) ++ans;
    }
  }
  cout << ans;
}

```