Problem: Given a string s, process q queries, each having one of the following forms:

  * 1 i c — Change the i-th character in the string to c. 
  * 2 l r y — Consider the substring of s starting at position l and ending at position r. Output the number of times y occurs as a substring in it. 

Input

The first line of the input contains the string s (1 ≤ |s| ≤ 105) of lowercase English letters.

The second line contains an integer q (1 ≤ q ≤ 105) — the number of queries to process.

The next q lines describe the queries and may have one of the following forms:

  * 1 i c (1 ≤ i ≤ |s|) 
  * 2 l r y (1 ≤ l ≤ r ≤ |s|) 



c is a lowercase English letter and y is a non-empty string consisting of only lowercase English letters.

The sum of |y| over all queries of second type is at most 105.

It is guaranteed that there is at least one query of second type.

All strings are 1-indexed.

|s| is the length of the string s.

Output

For each query of type 2, output the required answer in a separate line.

Examples

Input

ababababa
3
2 1 7 aba
1 5 c
2 1 7 aba


Output

3
1


Input

abcdcbc
5
2 1 7 bc
1 4 b
2 4 7 bc
1 2 a
2 1 4 aa


Output

2
2
1

Note

Consider the first sample case. Initially, the string aba occurs 3 times in the range [1, 7]. Note that two occurrences may overlap. 

After the update, the string becomes ababcbaba and now aba occurs only once in the range [1, 7].

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = (int)1e5;
string s;
int len;
int q;
bitset<N> pos[26];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> s >> q;
  len = (int)s.size();
  for (int i = 0; i < len; i++) {
    pos[s[i] - 'a'][i] = 1;
  }
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int i;
      string t;
      cin >> i >> t;
      i--;
      if (s[i] != t[0]) {
        pos[s[i] - 'a'][i] = 0;
        pos[t[0] - 'a'][i] = 1;
        s[i] = t[0];
      }
    } else {
      int l, r, sz;
      string t;
      cin >> l >> r >> t;
      sz = (int)t.size();
      l--;
      r--;
      r = r - sz + 1;
      if (l > r) {
        cout << "0\n";
        continue;
      }
      bitset<N> good;
      good.reset();
      good = ~good;
      for (int i = 0; i < sz; i++) {
        good &= (pos[t[i] - 'a'] >> i);
      }
      int ans = 0;
      good >>= l;
      ans = good.count();
      good >>= (r - l + 1);
      ans -= good.count();
      cout << ans << "\n";
    }
  }
}