#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000051373, N = 1e6 + 10, Base = 119;
int phash[N], S, T, ans, Basepower[N], o, z;
string s, t;
int sum(int a, int b) {
  int c = a + b;
  if (c >= MOD) c -= MOD;
  return c;
}
int mul(int a, int b) { return 1ll * a * b % MOD; }
bool isEqual(int l1, int r1, int l2, int r2) {
  int hash1, hash2;
  if (r1 - l1 + 1 != r2 - l2 + 1) return false;
  if (r1 > r2) {
    swap(l1, l2);
    swap(r1, r2);
  }
  if (l1 == 0)
    hash1 = phash[r1];
  else
    hash1 = sum(phash[r1], MOD - phash[l1 - 1]);
  if (l2 == 0)
    hash2 = phash[r2];
  else
    hash2 = sum(phash[r2], MOD - phash[l2 - 1]);
  hash1 = mul(hash1, Basepower[l2 - l1]);
  return (hash1 == hash2);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  cin >> t;
  S = (int)s.size();
  T = (int)t.size();
  phash[0] = (int)(t[0] - 'a');
  Basepower[0] = 1;
  for (int i = 1; i < T; i++) {
    Basepower[i] = mul(Basepower[i - 1], Base);
    phash[i] = sum(phash[i - 1], mul((int)t[i] - 'a', Basepower[i]));
  }
  for (int i = 0; i < S; i++) {
    if (s[i] == '1')
      o++;
    else
      z++;
  }
  if (o > z) {
    for (int i = 1; i < (T / o); i++) {
      int sizeZ = T - (i * o);
      if (sizeZ % z != 0) continue;
      sizeZ /= z;
      if (sizeZ == 0) continue;
      int cur = 0, sizeO = i, startZ = -1, endZ, startO = -1, endO;
      bool f = true;
      for (int j = 0; j < S && f; j++) {
        if (s[j] == '0') {
          if (startZ == -1) {
            startZ = cur;
            cur += sizeZ;
            endZ = cur - 1;
          } else {
            if (!isEqual(startZ, endZ, cur, cur + sizeZ - 1)) {
              f = false;
              break;
            }
            cur += sizeZ;
          }
        } else {
          if (startO == -1) {
            startO = cur;
            cur += sizeO;
            endO = cur - 1;
          } else {
            if (!isEqual(startO, endO, cur, cur + sizeO - 1)) {
              f = false;
              break;
            }
            cur += sizeO;
          }
        }
        if (!f) break;
      }
      if (f && !isEqual(startZ, endZ, startO, endO)) ans++;
    }
  } else {
    for (int i = 1; i < (T / z); i++) {
      int sizeO = T - (i * z);
      if (sizeO % o != 0) continue;
      sizeO /= o;
      if (sizeO == 0) continue;
      int cur = 0, sizeZ = i, startZ = -1, endZ, startO = -1, endO;
      bool f = true;
      for (int j = 0; j < S && f; j++) {
        if (s[j] == '0') {
          if (startZ == -1) {
            startZ = cur;
            cur += sizeZ;
            endZ = cur - 1;
          } else {
            if (!isEqual(startZ, endZ, cur, cur + sizeZ - 1)) {
              f = false;
              break;
            }
            cur += sizeZ;
          }
        } else {
          if (startO == -1) {
            startO = cur;
            cur += sizeO;
            endO = cur - 1;
          } else {
            if (!isEqual(startO, endO, cur, cur + sizeO - 1)) {
              f = false;
              break;
            }
            cur += sizeO;
          }
        }
        if (!f) break;
      }
      if (f && !isEqual(startZ, endZ, startO, endO)) {
        ans++;
      }
    }
  }
  cout << ans;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 1000051373, N = 1e6 + 10, Base = 119;
int phash[N], S, T, ans, Basepower[N], o, z;
string s, t;
int sum(int a, int b) {
  int c = a + b;
  if (c >= MOD) c -= MOD;
  return c;
}
int mul(int a, int b) { return 1ll * a * b % MOD; }
bool isEqual(int l1, int r1, int l2, int r2) {
  int hash1, hash2;
  if (r1 - l1 + 1 != r2 - l2 + 1) return false;
  if (r1 > r2) {
    swap(l1, l2);
    swap(r1, r2);
  }
  if (l1 == 0)
    hash1 = phash[r1];
  else
    hash1 = sum(phash[r1], MOD - phash[l1 - 1]);
  if (l2 == 0)
    hash2 = phash[r2];
  else
    hash2 = sum(phash[r2], MOD - phash[l2 - 1]);
  hash1 = mul(hash1, Basepower[l2 - l1]);
  return (hash1 == hash2);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  cin >> t;
  S = (int)s.size();
  T = (int)t.size();
  phash[0] = (int)(t[0] - 'a');
  Basepower[0] = 1;
  for (int i = 1; i < T; i++) {
    Basepower[i] = mul(Basepower[i - 1], Base);
    phash[i] = sum(phash[i - 1], mul((int)t[i] - 'a', Basepower[i]));
  }
  for (int i = 0; i < S; i++) {
    if (s[i] == '1')
      o++;
    else
      z++;
  }
  if (o > z) {
    for (int i = 1; i < (T / o); i++) {
      int sizeZ = T - (i * o);
      if (sizeZ % z != 0) continue;
      sizeZ /= z;
      if (sizeZ == 0) continue;
      int cur = 0, sizeO = i, startZ = -1, endZ, startO = -1, endO;
      bool f = true;
      for (int j = 0; j < S && f; j++) {
        if (s[j] == '0') {
          if (startZ == -1) {
            startZ = cur;
            cur += sizeZ;
            endZ = cur - 1;
          } else {
            if (!isEqual(startZ, endZ, cur, cur + sizeZ - 1)) {
              f = false;
              break;
            }
            cur += sizeZ;
          }
        } else {
          if (startO == -1) {
            startO = cur;
            cur += sizeO;
            endO = cur - 1;
          } else {
            if (!isEqual(startO, endO, cur, cur + sizeO - 1)) {
              f = false;
              break;
            }
            cur += sizeO;
          }
        }
        if (!f) break;
      }
      if (f && !isEqual(startZ, endZ, startO, endO)) ans++;
    }
  } else {
    for (int i = 1; i < (T / z); i++) {
      int sizeO = T - (i * z);
      if (sizeO % o != 0) continue;
      sizeO /= o;
      if (sizeO == 0) continue;
      int cur = 0, sizeZ = i, startZ = -1, endZ, startO = -1, endO;
      bool f = true;
      for (int j = 0; j < S && f; j++) {
        if (s[j] == '0') {
          if (startZ == -1) {
            startZ = cur;
            cur += sizeZ;
            endZ = cur - 1;
          } else {
            if (!isEqual(startZ, endZ, cur, cur + sizeZ - 1)) {
              f = false;
              break;
            }
            cur += sizeZ;
          }
        } else {
          if (startO == -1) {
            startO = cur;
            cur += sizeO;
            endO = cur - 1;
          } else {
            if (!isEqual(startO, endO, cur, cur + sizeO - 1)) {
              f = false;
              break;
            }
            cur += sizeO;
          }
        }
        if (!f) break;
      }
      if (f && !isEqual(startZ, endZ, startO, endO)) {
        ans++;
      }
    }
  }
  cout << ans;
}

```