#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 5, Mod = 1e9 + 7, MAXN = 1e5 + 10, LG = 20, M = 1e7,
          P = 727, Sq = 445 + 5;
const long long inf = 1e18 + 10;
int H[N], pw[N], cnt0, cnt1;
string s, t;
void Hash() {
  H[1] = t[0];
  for (int i = 2; i <= (int)t.size(); i++) {
    H[i] = (1ll * H[i - 1] * P % Mod + t[i - 1]) % Mod;
  }
}
int Get(int l, int r) {
  int ret = H[r];
  int res = 1ll * H[l - 1] * pw[r - l + 1] % Mod;
  ret = (ret - res + Mod) % Mod;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  pw[0] = 1;
  for (int i = 1; i < N; i++) {
    pw[i] = 1ll * pw[i - 1] * P % Mod;
  }
  cin >> s >> t;
  Hash();
  for (int i = 0; i < (int)s.size(); i++) {
    if (s[i] == '0') {
      cnt0++;
    } else {
      cnt1++;
    }
  }
  int sz = t.size(), ans = 0;
  for (int i = 1; i <= sz; i++) {
    long long op1 = 1ll * i * cnt0, op2 = sz - op1;
    if (op2 <= 0 || op2 % cnt1) {
      continue;
    }
    op2 /= cnt1;
    op1 = i;
    int rem = 0, x, y;
    bool s0 = false, s1 = false;
    bool ok = true;
    for (int j = 0; j < (int)s.size(); j++) {
      if (s[j] == '0') {
        if (!s0) {
          x = Get(rem + 1, rem + op1);
          s0 = true;
        } else {
          if (Get(rem + 1, rem + op1) != x) {
            ok = false;
            break;
          }
        }
      }
      if (s[j] == '1') {
        if (!s1) {
          y = Get(rem + 1, rem + op2);
          s1 = true;
        } else {
          if (Get(rem + 1, rem + op2) != y) {
            ok = false;
            break;
          }
        }
      }
      if (s[j] == '0') {
        rem += op1;
      } else {
        rem += op2;
      }
    }
    if (ok && (op1 != op2 || x != y)) {
      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 int N = 2e6 + 5, Mod = 1e9 + 7, MAXN = 1e5 + 10, LG = 20, M = 1e7,
          P = 727, Sq = 445 + 5;
const long long inf = 1e18 + 10;
int H[N], pw[N], cnt0, cnt1;
string s, t;
void Hash() {
  H[1] = t[0];
  for (int i = 2; i <= (int)t.size(); i++) {
    H[i] = (1ll * H[i - 1] * P % Mod + t[i - 1]) % Mod;
  }
}
int Get(int l, int r) {
  int ret = H[r];
  int res = 1ll * H[l - 1] * pw[r - l + 1] % Mod;
  ret = (ret - res + Mod) % Mod;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  pw[0] = 1;
  for (int i = 1; i < N; i++) {
    pw[i] = 1ll * pw[i - 1] * P % Mod;
  }
  cin >> s >> t;
  Hash();
  for (int i = 0; i < (int)s.size(); i++) {
    if (s[i] == '0') {
      cnt0++;
    } else {
      cnt1++;
    }
  }
  int sz = t.size(), ans = 0;
  for (int i = 1; i <= sz; i++) {
    long long op1 = 1ll * i * cnt0, op2 = sz - op1;
    if (op2 <= 0 || op2 % cnt1) {
      continue;
    }
    op2 /= cnt1;
    op1 = i;
    int rem = 0, x, y;
    bool s0 = false, s1 = false;
    bool ok = true;
    for (int j = 0; j < (int)s.size(); j++) {
      if (s[j] == '0') {
        if (!s0) {
          x = Get(rem + 1, rem + op1);
          s0 = true;
        } else {
          if (Get(rem + 1, rem + op1) != x) {
            ok = false;
            break;
          }
        }
      }
      if (s[j] == '1') {
        if (!s1) {
          y = Get(rem + 1, rem + op2);
          s1 = true;
        } else {
          if (Get(rem + 1, rem + op2) != y) {
            ok = false;
            break;
          }
        }
      }
      if (s[j] == '0') {
        rem += op1;
      } else {
        rem += op2;
      }
    }
    if (ok && (op1 != op2 || x != y)) {
      ans++;
    }
  }
  cout << ans;
  return 0;
}

```