#include <bits/stdc++.h>
using namespace std;
const int P = 131;
const int M = 1e9 + 7;
const int maxs = 1e5 + 5;
const int maxt = 1e6 + 5;
int lens, lent;
char s[maxs], t[maxt];
int tot0, tot1;
long long pre[maxt], Ppow[maxt];
void prework(int len) {
  pre[0] = 0;
  Ppow[0] = 1;
  for (int i = 1; i <= len; i++) {
    pre[i] = pre[i - 1] * P + (t[i] - 'a' + 1), pre[i] %= M;
    Ppow[i] = Ppow[i - 1] * P % M;
  }
}
inline long long Hash(int l, int r) {
  return (pre[r] - pre[l - 1] * Ppow[r - (l - 1)] % M + M) % M;
}
int main() {
  scanf("%s%s", s + 1, t + 1);
  lens = strlen(s + 1), lent = strlen(t + 1);
  tot0 = tot1 = 0;
  for (int i = 1; i <= lens; i++) tot0 += (s[i] == '0'), tot1 += (s[i] == '1');
  int ans = 0;
  int len0, len1;
  prework(lent);
  for (len0 = 1; len0 < lent; len0++) {
    if (tot0 * len0 >= lent) break;
    if ((lent - tot0 * len0) % tot1 > 0) continue;
    len1 = (lent - tot0 * len0) / tot1;
    bool ok = 1;
    int cnt0 = 0, cnt1 = 0;
    long long hash_r0 = 0, hash_r1 = 0;
    for (int i = 1; i <= lens; i++) {
      if (s[i] == '0') {
        int l = cnt0 * len0 + cnt1 * len1 + 1, r = l + len0 - 1;
        long long has = Hash(l, r);
        if (hash_r0 == 0) {
          hash_r0 = has;
        } else {
          if (hash_r0 != has) {
            ok = 0;
            break;
          }
        }
        cnt0++;
      }
      if (s[i] == '1') {
        int l = cnt0 * len0 + cnt1 * len1 + 1, r = l + len1 - 1;
        long long has = Hash(l, r);
        if (hash_r1 == 0) {
          hash_r1 = has;
        } else {
          if (hash_r1 != has) {
            ok = 0;
            break;
          }
        }
        cnt1++;
      }
      if (hash_r0 != 0 && hash_r1 != 0 && hash_r0 == hash_r1) {
        ok = 0;
        break;
      }
    }
    if (ok) ans++;
  }
  printf("%d\n", ans);
}

### Prompt

Your challenge is to write 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 P = 131;
const int M = 1e9 + 7;
const int maxs = 1e5 + 5;
const int maxt = 1e6 + 5;
int lens, lent;
char s[maxs], t[maxt];
int tot0, tot1;
long long pre[maxt], Ppow[maxt];
void prework(int len) {
  pre[0] = 0;
  Ppow[0] = 1;
  for (int i = 1; i <= len; i++) {
    pre[i] = pre[i - 1] * P + (t[i] - 'a' + 1), pre[i] %= M;
    Ppow[i] = Ppow[i - 1] * P % M;
  }
}
inline long long Hash(int l, int r) {
  return (pre[r] - pre[l - 1] * Ppow[r - (l - 1)] % M + M) % M;
}
int main() {
  scanf("%s%s", s + 1, t + 1);
  lens = strlen(s + 1), lent = strlen(t + 1);
  tot0 = tot1 = 0;
  for (int i = 1; i <= lens; i++) tot0 += (s[i] == '0'), tot1 += (s[i] == '1');
  int ans = 0;
  int len0, len1;
  prework(lent);
  for (len0 = 1; len0 < lent; len0++) {
    if (tot0 * len0 >= lent) break;
    if ((lent - tot0 * len0) % tot1 > 0) continue;
    len1 = (lent - tot0 * len0) / tot1;
    bool ok = 1;
    int cnt0 = 0, cnt1 = 0;
    long long hash_r0 = 0, hash_r1 = 0;
    for (int i = 1; i <= lens; i++) {
      if (s[i] == '0') {
        int l = cnt0 * len0 + cnt1 * len1 + 1, r = l + len0 - 1;
        long long has = Hash(l, r);
        if (hash_r0 == 0) {
          hash_r0 = has;
        } else {
          if (hash_r0 != has) {
            ok = 0;
            break;
          }
        }
        cnt0++;
      }
      if (s[i] == '1') {
        int l = cnt0 * len0 + cnt1 * len1 + 1, r = l + len1 - 1;
        long long has = Hash(l, r);
        if (hash_r1 == 0) {
          hash_r1 = has;
        } else {
          if (hash_r1 != has) {
            ok = 0;
            break;
          }
        }
        cnt1++;
      }
      if (hash_r0 != 0 && hash_r1 != 0 && hash_r0 == hash_r1) {
        ok = 0;
        break;
      }
    }
    if (ok) ans++;
  }
  printf("%d\n", ans);
}

```