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 = 100005;
char s[N], t[N];
bitset<N> bs[26], cur;
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  for (int i = 1; i <= n; i++) bs[s[i] - 'a'][i] = 1;
  int q;
  cin >> q;
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int idx;
      scanf("%d", &idx);
      scanf("%s", t);
      bs[s[idx] - 'a'][idx] = 0;
      s[idx] = t[0];
      bs[s[idx] - 'a'][idx] = 1;
    } else {
      int l, r;
      scanf("%d %d %s", &l, &r, t + 1);
      int m = strlen(t + 1);
      if (r - l + 1 < m) {
        printf("0\n");
        continue;
      }
      cur.reset();
      cur = ~cur;
      for (int i = 1; i <= m; i++) {
        cur &= (bs[t[i] - 'a'] >> (i - 1));
      }
      cur >>= l;
      int ans = cur.count();
      cur >>= (r - l - m + 2);
      ans -= cur.count();
      printf("%d\n", ans);
    }
  }
}