#include <bits/stdc++.h>
using namespace std;
const int mod[] = {1000000007, 1000000009};
const int base[] = {79, 271};
string s1, s2;
int pws[2][1000009], pre[2][1000009], lol[2], n, m;
pair<int, int> unini;
pair<int, int> getHash(int l, int r) {
  if (l == 0) {
    return make_pair(pre[0][r], pre[1][r]);
  }
  int i, j, t1, t2, t3, t4;
  long long int buf1, buf2, buf3, buf4;
  for (i = 0; i < 2; i++) {
    buf1 = (1ll * pre[i][l - 1] * pws[i][r - l + 1]) % mod[i];
    buf1 = (pre[i][r] - buf1) % mod[i];
    if (buf1 < 0) {
      buf1 += mod[i];
    }
    lol[i] = buf1;
  }
  return make_pair(lol[0], lol[1]);
}
int solve(int l1, int l2) {
  int i, j, t1, t2, t3, t4;
  pair<int, int> p1, p2;
  p1 = p2 = unini;
  t1 = 0;
  for (i = 0; i < n; i++) {
    if (s1[i] == '0') {
      if (p1 == unini) {
        p1 = getHash(t1, t1 + l1 - 1);
      } else {
        if (p1 != getHash(t1, t1 + l1 - 1)) {
          return 0;
        }
      }
      t1 += l1;
    } else {
      if (p2 == unini) {
        p2 = getHash(t1, t1 + l2 - 1);
      } else {
        if (p2 != getHash(t1, t1 + l2 - 1)) {
          return 0;
        }
      }
      t1 += l2;
    }
  }
  if (p1 == p2) {
    return 0;
  }
  return (t1 == m);
}
int main() {
  ios::sync_with_stdio(0);
  int i, j, t1, t2, t3, t4, cnt0, cnt1, ans = 0;
  string temp1, temp2;
  cin >> s1 >> s2;
  n = s1.length();
  m = s2.length();
  unini = make_pair(-1, -1);
  pws[0][0] = pws[1][0] = 1;
  for (i = 0; i < 2; i++) {
    for (j = 1; j <= m; j++) {
      pws[i][j] = (1ll * pws[i][j - 1] * base[i]) % mod[i];
    }
  }
  pre[0][0] = pre[1][0] = (s2[0] - 'a' + 1);
  for (i = 0; i < 2; i++) {
    for (j = 1; j < m; j++) {
      pre[i][j] = (1ll * pre[i][j - 1] * base[i]) % mod[i];
      pre[i][j] += (s2[j] - 'a' + 1);
      pre[i][j] %= mod[i];
    }
  }
  if (s1[0] == '1') {
    for (i = 0; i < n; i++) {
      if (s1[i] == '0') {
        s1[i] = '1';
      } else {
        s1[i] = '0';
      }
    }
  }
  cnt0 = cnt1 = 0;
  for (i = 0; i < n; i++) {
    if (s1[i] == '0') {
      cnt0++;
    } else {
      cnt1++;
    }
  }
  for (i = 1; i <= m; i++) {
    t1 = cnt0 * i;
    t2 = (m - t1) / cnt1;
    t3 = (cnt1 * t2) + t1;
    if ((m - t1) % cnt1 || t3 != m || t2 <= 0) {
      continue;
    }
    ans += solve(i, (m - t1) / cnt1);
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate 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 mod[] = {1000000007, 1000000009};
const int base[] = {79, 271};
string s1, s2;
int pws[2][1000009], pre[2][1000009], lol[2], n, m;
pair<int, int> unini;
pair<int, int> getHash(int l, int r) {
  if (l == 0) {
    return make_pair(pre[0][r], pre[1][r]);
  }
  int i, j, t1, t2, t3, t4;
  long long int buf1, buf2, buf3, buf4;
  for (i = 0; i < 2; i++) {
    buf1 = (1ll * pre[i][l - 1] * pws[i][r - l + 1]) % mod[i];
    buf1 = (pre[i][r] - buf1) % mod[i];
    if (buf1 < 0) {
      buf1 += mod[i];
    }
    lol[i] = buf1;
  }
  return make_pair(lol[0], lol[1]);
}
int solve(int l1, int l2) {
  int i, j, t1, t2, t3, t4;
  pair<int, int> p1, p2;
  p1 = p2 = unini;
  t1 = 0;
  for (i = 0; i < n; i++) {
    if (s1[i] == '0') {
      if (p1 == unini) {
        p1 = getHash(t1, t1 + l1 - 1);
      } else {
        if (p1 != getHash(t1, t1 + l1 - 1)) {
          return 0;
        }
      }
      t1 += l1;
    } else {
      if (p2 == unini) {
        p2 = getHash(t1, t1 + l2 - 1);
      } else {
        if (p2 != getHash(t1, t1 + l2 - 1)) {
          return 0;
        }
      }
      t1 += l2;
    }
  }
  if (p1 == p2) {
    return 0;
  }
  return (t1 == m);
}
int main() {
  ios::sync_with_stdio(0);
  int i, j, t1, t2, t3, t4, cnt0, cnt1, ans = 0;
  string temp1, temp2;
  cin >> s1 >> s2;
  n = s1.length();
  m = s2.length();
  unini = make_pair(-1, -1);
  pws[0][0] = pws[1][0] = 1;
  for (i = 0; i < 2; i++) {
    for (j = 1; j <= m; j++) {
      pws[i][j] = (1ll * pws[i][j - 1] * base[i]) % mod[i];
    }
  }
  pre[0][0] = pre[1][0] = (s2[0] - 'a' + 1);
  for (i = 0; i < 2; i++) {
    for (j = 1; j < m; j++) {
      pre[i][j] = (1ll * pre[i][j - 1] * base[i]) % mod[i];
      pre[i][j] += (s2[j] - 'a' + 1);
      pre[i][j] %= mod[i];
    }
  }
  if (s1[0] == '1') {
    for (i = 0; i < n; i++) {
      if (s1[i] == '0') {
        s1[i] = '1';
      } else {
        s1[i] = '0';
      }
    }
  }
  cnt0 = cnt1 = 0;
  for (i = 0; i < n; i++) {
    if (s1[i] == '0') {
      cnt0++;
    } else {
      cnt1++;
    }
  }
  for (i = 1; i <= m; i++) {
    t1 = cnt0 * i;
    t2 = (m - t1) / cnt1;
    t3 = (cnt1 * t2) + t1;
    if ((m - t1) % cnt1 || t3 != m || t2 <= 0) {
      continue;
    }
    ans += solve(i, (m - t1) / cnt1);
  }
  printf("%d\n", ans);
  return 0;
}

```