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 maxn = 1e5 + 5;
bitset<maxn> rec[30], ans;
char s1[maxn], s2[maxn];
int idx(char c) { return c - 'a'; }
int main() {
  scanf("%s", s1 + 1);
  int n = strlen(s1 + 1);
  for (int i = 1; i <= n; ++i) rec[idx(s1[i])][i] = 1;
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int pos;
      char ch;
      scanf("%d %c", &pos, &ch);
      rec[idx(s1[pos])][pos] = 0;
      rec[idx(ch)][pos] = 1;
      s1[pos] = ch;
    } else {
      int l, r;
      scanf("%d%d%s", &l, &r, s2);
      int m = strlen(s2);
      ans.set();
      for (int i = 0; i < m; ++i) ans &= (rec[idx(s2[i])] >> i);
      int cntl = (ans >> l).count();
      int cntr = (ans >> (r - m + 2)).count();
      if (cntl < cntr)
        puts("0");
      else
        printf("%d\n", cntl - cntr);
    }
  }
  return 0;
}