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 = 100100;
int m, o, l, r;
char s[N], t[N];
bitset<N> x[30], ans, one;
int main() {
  scanf("%s", s);
  int n = strlen(s);
  for (int i = 0; i < n; i++) {
    int ch = s[i] - 'a';
    x[ch][i + 1] = 1;
  }
  scanf("%d", &m);
  for (int i = 0; i < n; i++) {
    one[i] = 1;
  }
  for (int i = 0; i < m; i++) {
    scanf("%d", &o);
    if (o == 1) {
      scanf("%d %s", &l, t);
      l--;
      x[s[l] - 'a'][l + 1] = 0;
      s[l] = t[0];
      x[s[l] - 'a'][l + 1] = 1;
    } else if (o == 2) {
      scanf("%d %d %s", &l, &r, t);
      int len = strlen(t);
      if (r - l + 1 < len) {
        puts("0");
        continue;
      }
      ans = one;
      for (int j = 0; j < len; j++) {
        int ch = t[j] - 'a';
        ans &= (x[ch] >> (j + 1));
      }
      int w = (ans >> (l - 1)).count() - (ans >> (r - len + 1)).count();
      printf("%d\n", w);
    }
  }
  return 0;
}