#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
struct Hasher {
  vector<ll> power = {1}, pref = {0};
  ll mod, q;
  char c;
  Hasher(const string& s, ll m, ll _q, char _c) : mod(m), q(_q), c(_c) {
    for (char x : s) {
      power.push_back(power.back() * q % mod);
      pref.push_back((pref.back() * q % mod + (x - c)) % mod);
    }
  }
  ll hash(int l, int r) {
    return (pref[r] - power[r - l] * pref[l] % mod + mod) % mod;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  string a, b;
  cin >> a >> b;
  Hasher hash(b, 1700000009, 29, 'a');
  vector<ll> count(2);
  vector<ll> pos(2, 0x3FFFFFFFFFFll);
  for (ll i = 0; i < (int)(a).size(); i++) {
    count[a[i] - '0']++;
    pos[a[i] - '0'] = min(i, pos[a[i] - '0']);
  }
  ll res = 0;
  for (ll i = 1; ((int)(b).size() - count[0] * i) > 0; i++) {
    if (((int)(b).size() - count[0] * i) % count[1] == 0) {
      ll j = ((int)(b).size() - count[0] * i) / count[1];
      bool ok = true;
      vector<ll> length = {i, j};
      vector<ll> compPos = {j * pos[0], i * pos[1]};
      vector<ll> comp = {hash.hash(j * pos[0], j * pos[0] + length[0]),
                         hash.hash(i * pos[1], i * pos[1] + length[1])};
      if (comp[0] == comp[1] && i == j) continue;
      for (ll c = 0, d = 0; c < (int)(a).size() && ok; c++) {
        ll cur = a[c] - '0';
        ok = hash.hash(d, d + length[cur]) == comp[cur];
        d += length[cur];
      }
      if (ok) res++;
    }
  }
  cout << res << endl;
}

### Prompt

Please create a solution in Cpp 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;
using ll = long long;
using ld = long double;
struct Hasher {
  vector<ll> power = {1}, pref = {0};
  ll mod, q;
  char c;
  Hasher(const string& s, ll m, ll _q, char _c) : mod(m), q(_q), c(_c) {
    for (char x : s) {
      power.push_back(power.back() * q % mod);
      pref.push_back((pref.back() * q % mod + (x - c)) % mod);
    }
  }
  ll hash(int l, int r) {
    return (pref[r] - power[r - l] * pref[l] % mod + mod) % mod;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  string a, b;
  cin >> a >> b;
  Hasher hash(b, 1700000009, 29, 'a');
  vector<ll> count(2);
  vector<ll> pos(2, 0x3FFFFFFFFFFll);
  for (ll i = 0; i < (int)(a).size(); i++) {
    count[a[i] - '0']++;
    pos[a[i] - '0'] = min(i, pos[a[i] - '0']);
  }
  ll res = 0;
  for (ll i = 1; ((int)(b).size() - count[0] * i) > 0; i++) {
    if (((int)(b).size() - count[0] * i) % count[1] == 0) {
      ll j = ((int)(b).size() - count[0] * i) / count[1];
      bool ok = true;
      vector<ll> length = {i, j};
      vector<ll> compPos = {j * pos[0], i * pos[1]};
      vector<ll> comp = {hash.hash(j * pos[0], j * pos[0] + length[0]),
                         hash.hash(i * pos[1], i * pos[1] + length[1])};
      if (comp[0] == comp[1] && i == j) continue;
      for (ll c = 0, d = 0; c < (int)(a).size() && ok; c++) {
        ll cur = a[c] - '0';
        ok = hash.hash(d, d + length[cur]) == comp[cur];
        d += length[cur];
      }
      if (ok) res++;
    }
  }
  cout << res << endl;
}

```