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 + 5, B = 233;
char s[N], str[N];
bitset<N> ch[26];
int main() {
  scanf("%s", str);
  int n = (int)strlen(str);
  for (int i = 0; i < n; ++i) ch[str[i] - 'a'].set(i);
  int q;
  scanf("%d", &q);
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 2) {
      int l, r;
      scanf("%d%d%s", &l, &r, s);
      --l, --r;
      int m = (int)strlen(s);
      if (r - l + 1 < m) {
        puts("0");
        continue;
      }
      bitset<N> pos = ch[s[0] - 'a'];
      for (int i = 1; i < m; ++i) pos &= ch[s[i] - 'a'] >> i;
      printf("%d\n", (int)(pos >> l << (N - (r - l + 1 - m + 1))).count());
    } else {
      int x;
      scanf("%d%s", &x, s);
      --x;
      ch[str[x] - 'a'].reset(x);
      str[x] = s[0];
      ch[str[x] - 'a'].set(x);
    }
  }
}