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;
char str[100005];
bitset<100005> temp[26], ans;
int main() {
  int n;
  scanf("%s", str);
  scanf("%d", &n);
  for (int i = 0; str[i]; ++i) {
    if (str[i] >= 'a' && str[i] <= 'z') {
      temp[str[i] - 'a'][i + 1] = 1;
    }
  }
  for (int i = 0; i < n; ++i) {
    int op;
    scanf("%d", &op);
    switch (op) {
      case 1:
        int pos;
        char v;
        scanf("%d %c", &pos, &v);
        temp[str[pos - 1] - 'a'][pos] = 0;
        temp[v - 'a'][pos] = 1;
        str[pos - 1] = v;
        break;
      case 2:
        int tl, tr;
        char str2[100005] = {0};
        scanf("%d %d %s", &tl, &tr, str2);
        ans.set();
        int j;
        for (j = 0; str2[j]; ++j) {
          ans &= (temp[str2[j] - 'a'] >> j);
        }
        int left = (ans >> (tl)).count();
        int right = (ans >> (tr - j + 2)).count();
        if (left - right < 0) {
          left = right;
        }
        printf("%d\n", left - right);
        break;
    }
  }
}