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 sz = 1e5 + 10;
bitset<sz> s[26], an;
int main() {
  string ss;
  cin >> ss;
  int n = ss.size(), q;
  for (int a = 0; a < n; a++) s[ss[a] - 'a'][a] = 1;
  cin >> q;
  for (int a = 0; a < q; a++) {
    int ty;
    scanf("%d", &ty);
    if (ty == 1) {
      int p;
      char sy;
      scanf("%d", &p);
      p--;
      cin >> sy;
      for (int b = 0; b < 26; b++) {
        if (sy - 'a' == b)
          s[b][p] = 1;
        else
          s[b][p] = 0;
      }
    } else {
      int l, r;
      string s2;
      scanf("%d%d", &l, &r);
      l--;
      cin >> s2;
      int m = s2.size();
      an.set();
      for (int b = 0; b < m; b++) an &= (s[s2[b] - 'a'] >> b);
      an >>= l;
      an <<= l;
      an <<= (sz - r + (m - 1));
      printf("%d\n", an.count());
    }
  }
}