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;
bitset<100001> f[26], ans;
int q[300001], h, t;
char tem[300001], str[100001];
int n, m;
int main() {
  scanf("%s", str + 1);
  n = strlen(str + 1);
  for (int i = 1; i <= n; i++) f[str[i] - 'a'].set(i);
  int q;
  scanf("%d", &q);
  int opt, j, c, l, r;
  ans.set();
  for (int i = 1; i <= q; i++) {
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d", &j);
      char ch = getchar();
      while (ch < 'a' || ch > 'z') ch = getchar();
      f[str[j] - 'a'][j] = 0;
      f[ch - 'a'][j] = 1;
      str[j] = ch;
    } else {
      scanf("%d%d%s", &l, &r, tem + 1);
      int len = strlen(tem + 1);
      ans.set();
      for (int j = 1; j <= len; j++) ans &= (f[tem[j] - 'a'] >> (j - 1));
      printf(
          "%d\n",
          max(int((ans >> (l)).count() - (ans >> (r - len + 2)).count()), 0));
    }
  }
}