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 = 100500;
bitset<N> used[26];
int main() {
  ios_base::sync_with_stdio(false);
  string s;
  int q;
  cin >> s >> q;
  for (int i = 0; i < s.size(); i++) used[s[i] - 'a'][i] = true;
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int ind;
      char c;
      cin >> ind >> c;
      ind--;
      used[s[ind] - 'a'][ind] = false;
      used[c - 'a'][ind] = true;
      s[ind] = c;
    } else {
      int l, r;
      string s;
      cin >> l >> r >> s;
      l--, r--;
      if (r - l + 1 < s.size())
        cout << 0 << endl;
      else {
        bitset<N> ans;
        ans.set();
        for (int i = 0; i < s.size(); i++) ans &= used[s[i] - 'a'] >> i;
        cout << (ans >> l).count() - (ans >> (r - s.size() + 2)).count()
             << endl;
      }
    }
  }
  return 0;
}