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;
template <typename T>
void read(T& n) {
  char ch;
  int sign = 1;
  while (!isdigit(ch = getchar()))
    if (ch == '-') sign = -1;
  n = ch - '0';
  while (isdigit(ch = getchar())) n = n * 10 + ch - '0';
  n *= sign;
}
const int INF = 1e9 + 7;
const int N = 122222;
int n, m, l, r, first, len, opt;
char s[N], t[N];
bitset<N> occ[35], full;
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = (1); i <= (n); ++i) occ[s[i] - 'a'][i] = 1;
  for (int i = (0); i <= (N - 1); ++i) full[i] = 1;
  read(m);
  while (m--) {
    read(opt);
    if (opt == 1) {
      char ch;
      read(first), scanf("%c", &ch);
      occ[s[first] - 'a'][first] = 0, occ[ch - 'a'][first] = 1;
      s[first] = ch;
    } else {
      read(l), read(r), scanf("%s", t + 1);
      len = strlen(t + 1);
      if (r - l + 1 < len) {
        puts("0");
        continue;
      }
      bitset<N> cnt = full, tmp;
      for (int i = (1); i <= (len); ++i) cnt &= occ[t[i] - 'a'] >> i;
      tmp = cnt >> (r - len + 1), cnt >>= l - 1;
      printf("%d\n", (int)cnt.count() - (int)tmp.count());
    }
  }
  return 0;
}