#include <bits/stdc++.h>
using namespace std;
void io();
string strS, strT;
long long thash[1000010], p[1000010];
int first[2] = {INT_MAX, INT_MAX}, occ[2] = {0, 0};
int moreFreqDig;
long long gethsh(int l, int r) {
  if (l == 0) return thash[r];
  return (thash[r] - thash[l - 1] + 1000000009LL) % 1000000009LL;
}
int isGood(int l0, int l1) {
  int len[2] = {l0, l1};
  if (len[0] == 0 || len[1] == 0) return 0;
  if (len[0] * occ[0] + len[1] * occ[1] != strT.size()) return 0;
  long long hsh[2] = {
      gethsh(first[0] * len[1], first[0] * len[1] + len[0] - 1),
      gethsh(first[1] * len[0], first[1] * len[0] + len[1] - 1)};
  long long hshcheck0 =
      (p[strT.size() - 1 - first[0] * len[!0]] * hsh[0]) % 1000000009LL;
  long long hshcheck1 =
      (p[strT.size() - 1 - first[1] * len[!1]] * hsh[1]) % 1000000009LL;
  if (hshcheck0 == hshcheck1) return 0;
  int idx = 0;
  long long shash = 0;
  for (char dig : strS) {
    int which = dig - '0';
    shash = (shash + p[idx - first[which] * len[!which]] * hsh[which]) %
            1000000009LL;
    idx += len[which];
  }
  return shash == thash[strT.size() - 1];
}
void init() {
  cin >> strS >> strT;
  p[0] = 1;
  for (int i = 1; i < (int)(strT.size()); i++)
    p[i] = (p[i - 1] * 105943LL) % 1000000009LL;
  for (int i = 0; i < (int)(strT.size()); i++)
    thash[i] = ((long long)(strT[i]) * p[i]) % 1000000009LL;
  for (int i = 1; i < (int)(strT.size()); i++)
    thash[i] = (thash[i] + thash[i - 1]) % 1000000009LL;
  for (int i = 0; i < (int)(strS.size()); i++) {
    occ[strS[i] - '0']++;
    first[strS[i] - '0'] = min(first[strS[i] - '0'], i);
  }
  if (occ[0] > occ[1])
    moreFreqDig = 0;
  else
    moreFreqDig = 1;
}
int main() {
  io();
  init();
  int ans = 0;
  for (int len = 1; len <= (int)strT.size() / occ[moreFreqDig]; len++) {
    if (moreFreqDig == 0)
      ans += isGood(len, (strT.size() - occ[0] * len) / occ[1]);
    else
      ans += isGood((strT.size() - occ[1] * len) / occ[0], len);
  }
  cout << ans << "\n";
  return 0;
}
void io() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}

### Prompt

Create a solution in CPP for 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;
void io();
string strS, strT;
long long thash[1000010], p[1000010];
int first[2] = {INT_MAX, INT_MAX}, occ[2] = {0, 0};
int moreFreqDig;
long long gethsh(int l, int r) {
  if (l == 0) return thash[r];
  return (thash[r] - thash[l - 1] + 1000000009LL) % 1000000009LL;
}
int isGood(int l0, int l1) {
  int len[2] = {l0, l1};
  if (len[0] == 0 || len[1] == 0) return 0;
  if (len[0] * occ[0] + len[1] * occ[1] != strT.size()) return 0;
  long long hsh[2] = {
      gethsh(first[0] * len[1], first[0] * len[1] + len[0] - 1),
      gethsh(first[1] * len[0], first[1] * len[0] + len[1] - 1)};
  long long hshcheck0 =
      (p[strT.size() - 1 - first[0] * len[!0]] * hsh[0]) % 1000000009LL;
  long long hshcheck1 =
      (p[strT.size() - 1 - first[1] * len[!1]] * hsh[1]) % 1000000009LL;
  if (hshcheck0 == hshcheck1) return 0;
  int idx = 0;
  long long shash = 0;
  for (char dig : strS) {
    int which = dig - '0';
    shash = (shash + p[idx - first[which] * len[!which]] * hsh[which]) %
            1000000009LL;
    idx += len[which];
  }
  return shash == thash[strT.size() - 1];
}
void init() {
  cin >> strS >> strT;
  p[0] = 1;
  for (int i = 1; i < (int)(strT.size()); i++)
    p[i] = (p[i - 1] * 105943LL) % 1000000009LL;
  for (int i = 0; i < (int)(strT.size()); i++)
    thash[i] = ((long long)(strT[i]) * p[i]) % 1000000009LL;
  for (int i = 1; i < (int)(strT.size()); i++)
    thash[i] = (thash[i] + thash[i - 1]) % 1000000009LL;
  for (int i = 0; i < (int)(strS.size()); i++) {
    occ[strS[i] - '0']++;
    first[strS[i] - '0'] = min(first[strS[i] - '0'], i);
  }
  if (occ[0] > occ[1])
    moreFreqDig = 0;
  else
    moreFreqDig = 1;
}
int main() {
  io();
  init();
  int ans = 0;
  for (int len = 1; len <= (int)strT.size() / occ[moreFreqDig]; len++) {
    if (moreFreqDig == 0)
      ans += isGood(len, (strT.size() - occ[0] * len) / occ[1]);
    else
      ans += isGood((strT.size() - occ[1] * len) / occ[0], len);
  }
  cout << ans << "\n";
  return 0;
}
void io() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}

```