#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
const int MOD1 = 1000010101;
const long long prm = 109;
long long h[1000005];
long long modpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (a * res) % MOD;
    a = (a * a) % MOD;
    b = b >> 1;
  }
  return res;
}
long long modPos(long long a, long long mod = MOD) {
  if (a >= 0) return a % mod;
  return (a % mod + mod) % mod;
}
bool is_equal(pair<int, int> p1, pair<int, int> p2) {
  long long mul = modpow(prm, abs(p1.first - p2.first));
  if (p1.first >= p2.first)
    return modPos(h[p1.second] - h[p1.first]) ==
           modPos(modPos(h[p2.second] - h[p2.first]) * mul);
  else
    return modPos((mul * modPos(h[p1.second] - h[p1.first]))) ==
           modPos(h[p2.second] - h[p2.first]);
}
long long p_pow[1000005];
long long get_hash(int l, int r) {
  if (l == 0) return h[r];
  return modPos(h[r] - ((h[l - 1] * p_pow[r - l + 1]) % MOD));
}
int main() {
  string s, t;
  cin >> s >> t;
  int n = t.length();
  h[0] = t[0] - 'a' + 1;
  p_pow[0] = 1;
  for (int i = (1); i < (n + 1); i++) p_pow[i] = (p_pow[i - 1] * prm) % MOD;
  for (int i = (1); i < (n); i++)
    h[i] = ((h[i - 1] * prm) % MOD + (t[i] - 'a' + 1)) % MOD;
  int cnt[] = {0, 0};
  for (int i = 0; i < (s.length()); i++) cnt[s[i] - '0']++;
  if (cnt[0] < cnt[1]) {
    swap(cnt[0], cnt[1]);
    for (int i = 0; i < (s.length()); i++) s[i] = '1' - s[i] + '0';
  }
  int ans = 0;
  for (int i = 1; i * cnt[0] < n; i++) {
    if ((n - i * cnt[0]) % cnt[1] == 0) {
      int j = (n - i * cnt[0]) / cnt[1];
      int cur_pos = 0;
      long long h0 = -1;
      long long h1 = -1;
      bool match = true;
      for (auto &ch : s) {
        if (ch == '0') {
          long long h = get_hash(cur_pos, cur_pos + i - 1);
          cur_pos += i;
          if (h0 == -1)
            h0 = h;
          else if (h0 != h) {
            match = false;
            break;
          }
        }
        if (ch == '1') {
          long long h = get_hash(cur_pos, cur_pos + j - 1);
          cur_pos += j;
          if (h1 == -1)
            h1 = h;
          else if (h1 != h) {
            match = false;
            break;
          }
        }
      }
      if (match && h0 != h1) ans++;
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Your task is to create 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 long long MOD = 1000000007;
const int MOD1 = 1000010101;
const long long prm = 109;
long long h[1000005];
long long modpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (a * res) % MOD;
    a = (a * a) % MOD;
    b = b >> 1;
  }
  return res;
}
long long modPos(long long a, long long mod = MOD) {
  if (a >= 0) return a % mod;
  return (a % mod + mod) % mod;
}
bool is_equal(pair<int, int> p1, pair<int, int> p2) {
  long long mul = modpow(prm, abs(p1.first - p2.first));
  if (p1.first >= p2.first)
    return modPos(h[p1.second] - h[p1.first]) ==
           modPos(modPos(h[p2.second] - h[p2.first]) * mul);
  else
    return modPos((mul * modPos(h[p1.second] - h[p1.first]))) ==
           modPos(h[p2.second] - h[p2.first]);
}
long long p_pow[1000005];
long long get_hash(int l, int r) {
  if (l == 0) return h[r];
  return modPos(h[r] - ((h[l - 1] * p_pow[r - l + 1]) % MOD));
}
int main() {
  string s, t;
  cin >> s >> t;
  int n = t.length();
  h[0] = t[0] - 'a' + 1;
  p_pow[0] = 1;
  for (int i = (1); i < (n + 1); i++) p_pow[i] = (p_pow[i - 1] * prm) % MOD;
  for (int i = (1); i < (n); i++)
    h[i] = ((h[i - 1] * prm) % MOD + (t[i] - 'a' + 1)) % MOD;
  int cnt[] = {0, 0};
  for (int i = 0; i < (s.length()); i++) cnt[s[i] - '0']++;
  if (cnt[0] < cnt[1]) {
    swap(cnt[0], cnt[1]);
    for (int i = 0; i < (s.length()); i++) s[i] = '1' - s[i] + '0';
  }
  int ans = 0;
  for (int i = 1; i * cnt[0] < n; i++) {
    if ((n - i * cnt[0]) % cnt[1] == 0) {
      int j = (n - i * cnt[0]) / cnt[1];
      int cur_pos = 0;
      long long h0 = -1;
      long long h1 = -1;
      bool match = true;
      for (auto &ch : s) {
        if (ch == '0') {
          long long h = get_hash(cur_pos, cur_pos + i - 1);
          cur_pos += i;
          if (h0 == -1)
            h0 = h;
          else if (h0 != h) {
            match = false;
            break;
          }
        }
        if (ch == '1') {
          long long h = get_hash(cur_pos, cur_pos + j - 1);
          cur_pos += j;
          if (h1 == -1)
            h1 = h;
          else if (h1 != h) {
            match = false;
            break;
          }
        }
      }
      if (match && h0 != h1) ans++;
    }
  }
  cout << ans;
  return 0;
}

```