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 pii = std::pair<int, int>;
using pll = std::pair<long long, long long>;
const int maxn = 1e5 + 10;
std::bitset<maxn> mask[26];
int main() {
  std::ios::sync_with_stdio(false);
  std::string in;
  std::cin >> in;
  for (size_t i = 0; i < in.length(); i++) mask[in[i] - 'a'][i] = 1;
  int q;
  std::cin >> q;
  while (q--) {
    int c;
    std::cin >> c;
    if (c == 1) {
      int x;
      char ch;
      std::cin >> x >> ch;
      --x;
      mask[in[x] - 'a'][x] = 0;
      in[x] = ch;
      mask[ch - 'a'][x] = 1;
    } else {
      int l, r;
      std::string pat;
      std::cin >> l >> r >> pat;
      --l;
      if (r - l < (int)pat.length()) {
        std::cout << "0\n";
        continue;
      }
      auto test = std::bitset<maxn>{}.set();
      for (size_t i = 0; i < pat.length(); i++) test &= mask[pat[i] - 'a'] >> i;
      int ans = (test >> l).count() - (test >> (r - pat.length() + 1)).count();
      std::cout << ans << "\n";
    }
  }
  return 0;
}