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 B = 2;
const int C = 26;
const int N = 100005;
char v[B];
char s[N], now[N];
int n, q, x, y, o;
std::bitset<N> a;
std::bitset<N> se[C];
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0), std::cout.tie(0);
  std::cin >> s + 1, n = strlen(s + 1);
  for (int i = 1; i <= n; ++i) se[s[i] - 'a'].set(i, 1);
  std::cin >> q;
  for (int i = 1; i <= q; ++i) {
    std::cin >> o;
    if (o == 1) {
      std::cin >> x >> v;
      se[s[x] - 'a'].set(x, 0), se[v[0] - 'a'].set(x, 1);
      s[x] = v[0];
    } else {
      std::cin >> x >> y >> now + 1;
      a.set();
      int l = strlen(now + 1);
      for (int i = 1; i <= l; ++i) a &= (se[now[i] - 'a'] >> (i - 1));
      int la = (a >> x).count(), lb = (a >> (y - l + 2)).count();
      std::cout << ((la - lb < 0) ? 0 : la - lb) << '\n';
    }
  }
  return 0;
}