#include <bits/stdc++.h>
using namespace std;
const int MOD1 = 1e9 + 7;
const int MOD2 = 1e9 + 9;
const int INF = 2e9 + 5;
const int SEED1 = 2687;
const int SEED2 = 4007;
string s, t;
int o = 0, z = 0;
long long hash1[1000010];
long long hash2[1000010];
int modpow(int b, int e, int mod) {
  if (e == 0) return 1;
  long long x = modpow(b, e / 2, mod);
  (x *= x) %= mod;
  if (e % 2 == 1) (x *= b) %= mod;
  return x;
}
long long inv(int x, int mod) { return modpow(x, mod - 2, mod); }
int value1(int idx, int nxt) {
  long long v1 = hash1[nxt];
  long long lo = (idx > 0 ? hash1[idx - 1] : 0);
  long long K = modpow(SEED1, nxt - idx + 1, MOD1);
  return (v1 - lo * K % MOD1 + MOD1) % MOD1;
}
int value2(int idx, int nxt) {
  long long v2 = hash2[nxt];
  long long lo = (idx > 0 ? hash2[idx - 1] : 0);
  long long K = modpow(SEED2, nxt - idx + 1, MOD2);
  return (v2 - lo * K % MOD2 + MOD2) % MOD2;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> s >> t;
  for (int i = 0; i < t.length(); i++) {
    int prev1 = i > 0 ? hash1[i - 1] : 0;
    int prev2 = i > 0 ? hash2[i - 1] : 0;
    hash1[i] = ((long long)prev1 * SEED1 + t[i]) % MOD1;
    hash2[i] = ((long long)prev2 * SEED2 + t[i]) % MOD2;
  }
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '0')
      ++z;
    else
      ++o;
  }
  int m = t.length();
  int possible = 0;
  for (int i = 1; i <= t.length(); i++) {
    long long v = m - (long long)i * o;
    if (v <= 0 || v % z != 0) continue;
    int j = v / z;
    int idx = 0;
    int zHash1 = -1, zHash2 = -1;
    int oHash1 = -1, oHash2 = -1;
    bool good = true;
    for (int k = 0; k < s.length(); k++) {
      if (s[k] == '0') {
        int v1 = value1(idx, idx + j - 1);
        int v2 = value2(idx, idx + j - 1);
        if (zHash1 == -1) {
          zHash1 = v1;
          zHash2 = v2;
        } else {
          if (v1 != zHash1 || v2 != zHash2) good = false;
        }
        idx += j;
      } else {
        int v1 = value1(idx, idx + i - 1);
        int v2 = value2(idx, idx + i - 1);
        if (oHash1 == -1) {
          oHash1 = v1;
          oHash2 = v2;
        } else {
          if (v1 != oHash1 || v2 != oHash2) good = false;
        }
        idx += i;
      }
    }
    if (zHash1 == oHash1 && zHash2 == oHash2) continue;
    if (good) ++possible;
  }
  cout << possible;
}

### Prompt

Please formulate 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 MOD1 = 1e9 + 7;
const int MOD2 = 1e9 + 9;
const int INF = 2e9 + 5;
const int SEED1 = 2687;
const int SEED2 = 4007;
string s, t;
int o = 0, z = 0;
long long hash1[1000010];
long long hash2[1000010];
int modpow(int b, int e, int mod) {
  if (e == 0) return 1;
  long long x = modpow(b, e / 2, mod);
  (x *= x) %= mod;
  if (e % 2 == 1) (x *= b) %= mod;
  return x;
}
long long inv(int x, int mod) { return modpow(x, mod - 2, mod); }
int value1(int idx, int nxt) {
  long long v1 = hash1[nxt];
  long long lo = (idx > 0 ? hash1[idx - 1] : 0);
  long long K = modpow(SEED1, nxt - idx + 1, MOD1);
  return (v1 - lo * K % MOD1 + MOD1) % MOD1;
}
int value2(int idx, int nxt) {
  long long v2 = hash2[nxt];
  long long lo = (idx > 0 ? hash2[idx - 1] : 0);
  long long K = modpow(SEED2, nxt - idx + 1, MOD2);
  return (v2 - lo * K % MOD2 + MOD2) % MOD2;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> s >> t;
  for (int i = 0; i < t.length(); i++) {
    int prev1 = i > 0 ? hash1[i - 1] : 0;
    int prev2 = i > 0 ? hash2[i - 1] : 0;
    hash1[i] = ((long long)prev1 * SEED1 + t[i]) % MOD1;
    hash2[i] = ((long long)prev2 * SEED2 + t[i]) % MOD2;
  }
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '0')
      ++z;
    else
      ++o;
  }
  int m = t.length();
  int possible = 0;
  for (int i = 1; i <= t.length(); i++) {
    long long v = m - (long long)i * o;
    if (v <= 0 || v % z != 0) continue;
    int j = v / z;
    int idx = 0;
    int zHash1 = -1, zHash2 = -1;
    int oHash1 = -1, oHash2 = -1;
    bool good = true;
    for (int k = 0; k < s.length(); k++) {
      if (s[k] == '0') {
        int v1 = value1(idx, idx + j - 1);
        int v2 = value2(idx, idx + j - 1);
        if (zHash1 == -1) {
          zHash1 = v1;
          zHash2 = v2;
        } else {
          if (v1 != zHash1 || v2 != zHash2) good = false;
        }
        idx += j;
      } else {
        int v1 = value1(idx, idx + i - 1);
        int v2 = value2(idx, idx + i - 1);
        if (oHash1 == -1) {
          oHash1 = v1;
          oHash2 = v2;
        } else {
          if (v1 != oHash1 || v2 != oHash2) good = false;
        }
        idx += i;
      }
    }
    if (zHash1 == oHash1 && zHash2 == oHash2) continue;
    if (good) ++possible;
  }
  cout << possible;
}

```