#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll maxn = 1e6 + 6, mod = 1e9 + 7;
ll n, m, ps[maxn], ans, n0, n1, p0 = -1, p1 = -1, pw[maxn];
string s, t;
bool check(ll p, ll p2, ll val0, ll val1, ll k0, ll k1) {
  if (p == n - 1) return 1;
  if (p >= n) return 0;
  if (s[p + 1] == '0') {
    ll val = (ps[p2 + k0 + 1] - ps[p2 + 1] + mod * 10) % mod;
    ll y = p2 + 1;
    y = pw[y];
    if (val != val0 * y % mod) {
      return 0;
    }
    return check(p + 1, p2 + k0, val0, val1, k0, k1);
  }
  ll y = p2 - p1 * k0 + 1;
  y = pw[y];
  ll val = (ps[p2 + k1 + 1] - ps[p2 + 1] + mod * 10) % mod;
  if (val != val1 * y % mod) return 0;
  return check(p + 1, p2 + k1, val0, val1, k0, k1);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  pw[0] = 1;
  for (ll i = 1; i < maxn; i++) pw[i] = (pw[i - 1] * 26) % mod;
  cin >> s >> t;
  n = s.size();
  m = t.size();
  for (ll i = 0; i < m; i++) {
    if (t[i] == 'b')
      t[i] = 'x';
    else if (t[i] == 'x')
      t[i] = 'b';
  }
  if (s[0] == '1') {
    for (ll i = 0; i < n; i++) s[i] = char('0' + 1 - (s[i] - '0'));
  }
  for (ll i = 0; i < n; i++) {
    if (s[i] == '0') {
      n0++;
      if (p0 == -1) p0 = i;
    } else {
      n1++;
      if (p1 == -1) p1 = i;
    }
  }
  for (ll i = 0; i < m; i++) {
    ll x = t[i] - 'a';
    ps[i + 1] = (ps[i] + pw[i] * x) % mod;
  }
  for (ll i = 0; i < m; i++) {
    ll x = i + 1;
    if (x * n0 + n1 > m) break;
    ll y = m - x * n0;
    if ((y <= 0) || (y % n1 > 0)) continue;
    y /= n1;
    ll val0, val1;
    ll z = pw[p1 * x];
    val0 = ps[x];
    val1 = (ps[p1 * x + y] - ps[p1 * x] + mod * 10) % mod;
    bool b = check(0, i, val0, val1, x, y);
    if (!b) continue;
    if (val1 == val0 * z % mod && x == y) continue;
    ans++;
  }
  cout << ans << endl;
}

### 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;
using ll = long long;
const ll maxn = 1e6 + 6, mod = 1e9 + 7;
ll n, m, ps[maxn], ans, n0, n1, p0 = -1, p1 = -1, pw[maxn];
string s, t;
bool check(ll p, ll p2, ll val0, ll val1, ll k0, ll k1) {
  if (p == n - 1) return 1;
  if (p >= n) return 0;
  if (s[p + 1] == '0') {
    ll val = (ps[p2 + k0 + 1] - ps[p2 + 1] + mod * 10) % mod;
    ll y = p2 + 1;
    y = pw[y];
    if (val != val0 * y % mod) {
      return 0;
    }
    return check(p + 1, p2 + k0, val0, val1, k0, k1);
  }
  ll y = p2 - p1 * k0 + 1;
  y = pw[y];
  ll val = (ps[p2 + k1 + 1] - ps[p2 + 1] + mod * 10) % mod;
  if (val != val1 * y % mod) return 0;
  return check(p + 1, p2 + k1, val0, val1, k0, k1);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  pw[0] = 1;
  for (ll i = 1; i < maxn; i++) pw[i] = (pw[i - 1] * 26) % mod;
  cin >> s >> t;
  n = s.size();
  m = t.size();
  for (ll i = 0; i < m; i++) {
    if (t[i] == 'b')
      t[i] = 'x';
    else if (t[i] == 'x')
      t[i] = 'b';
  }
  if (s[0] == '1') {
    for (ll i = 0; i < n; i++) s[i] = char('0' + 1 - (s[i] - '0'));
  }
  for (ll i = 0; i < n; i++) {
    if (s[i] == '0') {
      n0++;
      if (p0 == -1) p0 = i;
    } else {
      n1++;
      if (p1 == -1) p1 = i;
    }
  }
  for (ll i = 0; i < m; i++) {
    ll x = t[i] - 'a';
    ps[i + 1] = (ps[i] + pw[i] * x) % mod;
  }
  for (ll i = 0; i < m; i++) {
    ll x = i + 1;
    if (x * n0 + n1 > m) break;
    ll y = m - x * n0;
    if ((y <= 0) || (y % n1 > 0)) continue;
    y /= n1;
    ll val0, val1;
    ll z = pw[p1 * x];
    val0 = ps[x];
    val1 = (ps[p1 * x + y] - ps[p1 * x] + mod * 10) % mod;
    bool b = check(0, i, val0, val1, x, y);
    if (!b) continue;
    if (val1 == val0 * z % mod && x == y) continue;
    ans++;
  }
  cout << ans << endl;
}

```