#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e6 + 10;
const unsigned long long HAP[2] = {567, 568};
unsigned long long ts[2][2], fac[2][maxN], Ha[2][maxN];
char s[maxN], t[maxN];
int ans, cnt[2];
bool check(int op, int l, int r) {
  bool pd = true;
  for (int j = 0; j < 2; ++j)
    pd &= ts[op][j] == Ha[j][r] - Ha[j][l - 1] * fac[j][r - l + 1];
  return pd;
}
void HashInit(int n) {
  for (int j = 0; j < 2; ++j) {
    fac[j][0] = 1;
    for (int i = 1; i <= n; ++i) fac[j][i] = fac[j][i - 1] * HAP[j];
    for (int i = 1; i <= n; ++i) Ha[j][i] = Ha[j][i - 1] * HAP[j] + t[i];
  }
}
int main() {
  scanf("%s%s", s + 1, t + 1);
  int ls = strlen(s + 1), lt = strlen(t + 1);
  for (int i = 1; i <= ls; ++i) ++cnt[s[i] - '0'];
  HashInit(lt);
  for (int l0 = 1, l1; l0 * cnt[0] < lt; ++l0)
    if ((lt - l0 * cnt[0]) % cnt[1] == 0) {
      l1 = (lt - l0 * cnt[0]) / cnt[1];
      bool p0 = 1, p1 = 1, p = 1;
      int k = 1;
      for (int i = 1; i <= ls; ++i) {
        if (!p) break;
        if (s[i] == '0') {
          if (p0) {
            p0 = 0;
            for (int j = 0; j < 2; ++j)
              ts[0][j] = Ha[j][k + l0 - 1] - Ha[j][k - 1] * fac[j][l0];
            if (!p1 && l0 == l1) p = !check(1, k, k + l0 - 1);
          } else
            p = check(0, k, k + l0 - 1);
          k += l0;
        } else {
          if (p1) {
            p1 = 0;
            for (int j = 0; j < 2; ++j)
              ts[1][j] = Ha[j][k + l1 - 1] - Ha[j][k - 1] * fac[j][l1];
            if (!p0 && l0 == l1) p = !check(0, k, k + l1 - 1);
          } else
            p = check(1, k, k + l1 - 1);
          k += l1;
        }
      }
      if (p) ++ans;
    }
  printf("%d", ans);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 maxN = 1e6 + 10;
const unsigned long long HAP[2] = {567, 568};
unsigned long long ts[2][2], fac[2][maxN], Ha[2][maxN];
char s[maxN], t[maxN];
int ans, cnt[2];
bool check(int op, int l, int r) {
  bool pd = true;
  for (int j = 0; j < 2; ++j)
    pd &= ts[op][j] == Ha[j][r] - Ha[j][l - 1] * fac[j][r - l + 1];
  return pd;
}
void HashInit(int n) {
  for (int j = 0; j < 2; ++j) {
    fac[j][0] = 1;
    for (int i = 1; i <= n; ++i) fac[j][i] = fac[j][i - 1] * HAP[j];
    for (int i = 1; i <= n; ++i) Ha[j][i] = Ha[j][i - 1] * HAP[j] + t[i];
  }
}
int main() {
  scanf("%s%s", s + 1, t + 1);
  int ls = strlen(s + 1), lt = strlen(t + 1);
  for (int i = 1; i <= ls; ++i) ++cnt[s[i] - '0'];
  HashInit(lt);
  for (int l0 = 1, l1; l0 * cnt[0] < lt; ++l0)
    if ((lt - l0 * cnt[0]) % cnt[1] == 0) {
      l1 = (lt - l0 * cnt[0]) / cnt[1];
      bool p0 = 1, p1 = 1, p = 1;
      int k = 1;
      for (int i = 1; i <= ls; ++i) {
        if (!p) break;
        if (s[i] == '0') {
          if (p0) {
            p0 = 0;
            for (int j = 0; j < 2; ++j)
              ts[0][j] = Ha[j][k + l0 - 1] - Ha[j][k - 1] * fac[j][l0];
            if (!p1 && l0 == l1) p = !check(1, k, k + l0 - 1);
          } else
            p = check(0, k, k + l0 - 1);
          k += l0;
        } else {
          if (p1) {
            p1 = 0;
            for (int j = 0; j < 2; ++j)
              ts[1][j] = Ha[j][k + l1 - 1] - Ha[j][k - 1] * fac[j][l1];
            if (!p0 && l0 == l1) p = !check(0, k, k + l1 - 1);
          } else
            p = check(1, k, k + l1 - 1);
          k += l1;
        }
      }
      if (p) ++ans;
    }
  printf("%d", ans);
  return 0;
}

```