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 mod = 1e9 + 7;
const int maxn = 1e5 + 5;
bitset<maxn> f[26], w;
char s[maxn], p[maxn];
int main() {
  int n, q, op, i, j, k;
  scanf("%s", s);
  n = strlen(s);
  for (i = 0; i < n; i++) f[s[i] - 'a'][i] = 1;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d %s", &i, p);
      i--;
      f[s[i] - 'a'][i] = 0;
      s[i] = p[0];
      f[s[i] - 'a'][i] = 1;
    } else {
      scanf("%d%d %s", &i, &j, p);
      n = strlen(p);
      if (i + n - 1 > j) {
        puts("0");
        continue;
      }
      w = f[p[0] - 'a'];
      for (k = 1; k < n; k++) w &= f[p[k] - 'a'] >> k;
      printf("%d\n",
             (int)((w >> (i - 1)).count() - (w >> (j - n + 1)).count()));
    }
  }
  return 0;
}