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 = 1e5 + 1;
char c;
int n, m, t, x, y, l;
string s;
bitset<N> a[26], C, A;
int main() {
  cin >> s >> m, n = s.size();
  for (int i = 0; i < n; i++) a[s[i] - 'a'][i] = 1, C[i] = 1;
  while (m-- && cin >> t >> x)
    if (t == 1 && cin >> c)
      for (int i = 0; i < 26; i++) a[i][x - 1] = (c - 'a' == i);
    else {
      cin >> y >> s, l = s.size(), A = (C >> (n - y + x + l - 2)) << (x - 1);
      for (int i = 0; i < l; i++) A &= a[s[i] - 'a'], A <<= 1;
      cout << A.count() << '\n';
    }
  return 0;
}