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>
const int maxn = 100100;
std::bitset<maxn> s[26];
char ch[maxn], buf[maxn];
int n, m;
int main() {
  std::ios::sync_with_stdio(false), std::cin.tie(0);
  std::cin >> ch, n = strlen(ch);
  for (int i = 0; i < n; ++i) s[ch[i] - 'a'].set(i);
  for (std::cin >> m; m--;) {
    int opt, i, l, r;
    char c;
    std::cin >> opt;
    if (opt == 1) {
      std::cin >> i >> c, --i;
      s[ch[i] - 'a'].reset(i), s[(ch[i] = c) - 'a'].set(i);
    } else {
      std::cin >> l >> r >> buf;
      int len = strlen(buf);
      const int p = r - l + 1 - len + 1;
      if (p < 0) {
        std::cout << 0 << '\n';
        continue;
      }
      std::bitset<maxn> ans;
      ans.set();
      for (int i = 0; i < len; ++i) ans &= s[buf[i] - 'a'] >> i + l - 1;
      std::cout << ans.count() - (ans >> p).count() << '\n';
    }
  }
}