#include <bits/stdc++.h>
using namespace std;
const int MD = 1000000007;
const int MX = 1000005;
const long long R = 12345;
long long mod_mult(long long a, long long b) { return (a * b) % MD; }
long long rpow[MX];
long long hsh[MX];
long long gethash(int l, int r) {
  return (hsh[r] - (hsh[l - 1] * rpow[r - l + 1]) % MD + MD) % MD;
}
bool compare(int i, int j, int len) {
  return gethash(i, i + len - 1) == gethash(j, j + len - 1);
}
int pr[MX][2];
bool validlen[MX];
int main() {
  rpow[0] = 1;
  for (int i = 1; i < MX; i++) {
    rpow[i] = mod_mult(rpow[i - 1], R);
  }
  string s, t;
  cin >> s >> t;
  int n = s.size();
  int m = t.size();
  for (int i = 1; i <= m; i++) {
    hsh[i] = (hsh[i - 1] * R + t[i - 1] - 'a' + 195) % MD;
  }
  vector<int> c0pos, c1pos;
  for (int i = 1; i <= n; i++) {
    pr[i][0] = pr[i - 1][0];
    pr[i][1] = pr[i - 1][1];
    if (s[i - 1] == '0') {
      pr[i][0]++;
      c0pos.push_back(i);
    } else {
      pr[i][1]++;
      c1pos.push_back(i);
    }
  }
  int c0 = c0pos.size(), c1 = c1pos.size();
  for (int x = 1; x * c0 < m; x++) {
    int r = m - x * c0;
    if (r % c1 > 0) continue;
    int y = r / c1;
    bool ok = true;
    int ref = c0pos[0];
    ref = x * pr[ref - 1][0] + y * pr[ref - 1][1] + 1;
    for (int pos : c0pos) {
      pos = x * pr[pos - 1][0] + y * pr[pos - 1][1] + 1;
      if (!compare(ref, pos, x)) {
        ok = false;
        break;
      }
    }
    if (x == y) {
      int pos = 1;
      while (s[pos - 1] != '1') pos++;
      pos = x * pr[pos - 1][0] + y * pr[pos - 1][1] + 1;
      if (compare(ref, pos, x)) {
        ok = false;
      }
    }
    if (ok) {
      validlen[y] = true;
    }
  }
  int ans = 0;
  for (int y = 1; y * c1 < m; y++) {
    if (!validlen[y]) continue;
    int r = m - y * c1;
    if (r % c0 > 0) continue;
    int x = r / c0;
    int ref = c1pos[0];
    ref = x * pr[ref - 1][0] + y * pr[ref - 1][1] + 1;
    bool ok = true;
    for (int pos : c1pos) {
      pos = x * pr[pos - 1][0] + y * pr[pos - 1][1] + 1;
      if (!compare(ref, pos, y)) {
        ok = false;
        break;
      }
    }
    if (ok) {
      ans++;
    }
  }
  printf("%d\n", ans);
}

### 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 int MD = 1000000007;
const int MX = 1000005;
const long long R = 12345;
long long mod_mult(long long a, long long b) { return (a * b) % MD; }
long long rpow[MX];
long long hsh[MX];
long long gethash(int l, int r) {
  return (hsh[r] - (hsh[l - 1] * rpow[r - l + 1]) % MD + MD) % MD;
}
bool compare(int i, int j, int len) {
  return gethash(i, i + len - 1) == gethash(j, j + len - 1);
}
int pr[MX][2];
bool validlen[MX];
int main() {
  rpow[0] = 1;
  for (int i = 1; i < MX; i++) {
    rpow[i] = mod_mult(rpow[i - 1], R);
  }
  string s, t;
  cin >> s >> t;
  int n = s.size();
  int m = t.size();
  for (int i = 1; i <= m; i++) {
    hsh[i] = (hsh[i - 1] * R + t[i - 1] - 'a' + 195) % MD;
  }
  vector<int> c0pos, c1pos;
  for (int i = 1; i <= n; i++) {
    pr[i][0] = pr[i - 1][0];
    pr[i][1] = pr[i - 1][1];
    if (s[i - 1] == '0') {
      pr[i][0]++;
      c0pos.push_back(i);
    } else {
      pr[i][1]++;
      c1pos.push_back(i);
    }
  }
  int c0 = c0pos.size(), c1 = c1pos.size();
  for (int x = 1; x * c0 < m; x++) {
    int r = m - x * c0;
    if (r % c1 > 0) continue;
    int y = r / c1;
    bool ok = true;
    int ref = c0pos[0];
    ref = x * pr[ref - 1][0] + y * pr[ref - 1][1] + 1;
    for (int pos : c0pos) {
      pos = x * pr[pos - 1][0] + y * pr[pos - 1][1] + 1;
      if (!compare(ref, pos, x)) {
        ok = false;
        break;
      }
    }
    if (x == y) {
      int pos = 1;
      while (s[pos - 1] != '1') pos++;
      pos = x * pr[pos - 1][0] + y * pr[pos - 1][1] + 1;
      if (compare(ref, pos, x)) {
        ok = false;
      }
    }
    if (ok) {
      validlen[y] = true;
    }
  }
  int ans = 0;
  for (int y = 1; y * c1 < m; y++) {
    if (!validlen[y]) continue;
    int r = m - y * c1;
    if (r % c0 > 0) continue;
    int x = r / c0;
    int ref = c1pos[0];
    ref = x * pr[ref - 1][0] + y * pr[ref - 1][1] + 1;
    bool ok = true;
    for (int pos : c1pos) {
      pos = x * pr[pos - 1][0] + y * pr[pos - 1][1] + 1;
      if (!compare(ref, pos, y)) {
        ok = false;
        break;
      }
    }
    if (ok) {
      ans++;
    }
  }
  printf("%d\n", ans);
}

```