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 EXP = 26;
const int DIM = 100005;
char str[DIM], aux[DIM];
bitset<DIM> bts[EXP], sol, one;
int main(void) {
  scanf("%s", str + 1);
  int n = (int)strlen(str + 1);
  for (int i = 1; i <= n; ++i) {
    bts[str[i] - 'a'][i] = true;
    one[i - 1] = true;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int p;
      char ch;
      scanf("%d %c", &p, &ch);
      bts[str[p] - 'a'][p] = false;
      bts[ch - 'a'][p] = true;
      str[p] = ch;
    } else {
      int l, r;
      scanf("%d %d %s", &l, &r, aux + 1);
      int m = (int)strlen(aux + 1);
      if (m > r - l + 1)
        printf("%d\n", 0);
      else {
        sol = one;
        for (int i = 1; i <= m; ++i) sol &= (bts[aux[i] - 'a'] >> i);
        printf("%d\n", (int)(sol >> (l - 1)).count() -
                           (int)(sol >> (r - m + 1)).count());
      }
    }
  }
  return 0;
}