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 = 100000 + 5;
char s[N];
int q;
char y[N], c[3];
bitset<N> state[26], ans;
int main() {
  scanf("%s%d", (s + 1), &q);
  int n = strlen(s + 1);
  for (register int i = 1; i <= n; i++) state[s[i] - 'a'].set(i);
  while (q-- > 0) {
    int op, l, r;
    scanf("%d%d", &op, &l);
    if (op == 1) {
      scanf("%s", c);
      char rw = c[0];
      int x = s[l] - 'a', y = rw - 'a';
      state[x].reset(l);
      state[y].set(l);
      s[l] = rw;
    } else if (op == 2) {
      scanf("%d%s", &r, y);
      ans.set();
      int v = strlen(y);
      r = r - v + 1;
      for (int i = 0; i < v; i++) ans &= (state[y[i] - 'a'] >> i);
      ans <<= (N - r - 1);
      ans >>= (N - r - 1);
      ans >>= l;
      printf("%d\n", ans.count());
    }
  }
  return 0;
}