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;
namespace Primary {
const int maxn = 1e5 + 10;
char s[maxn], t[maxn];
int q;
bitset<maxn> help[26], ans;
void main() {
  scanf("%s%d", s, &q);
  int len = strlen(s);
  for (int i = 0; i < len; ++i) help[s[i] - 'a'].set(i);
  int opt, ql, qr;
  while (q--) {
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d%s", &ql, t), --ql;
      help[s[ql] - 'a'].set(ql, 0);
      help[t[0] - 'a'].set(ql);
      s[ql] = t[0];
    } else {
      scanf("%d%d%s", &ql, &qr, t);
      len = strlen(t);
      if (len > qr - ql + 1) {
        puts("0");
        continue;
      }
      ans.set();
      for (int i = 0; i < len; ++i) ans &= (help[t[i] - 'a'] >> i);
      int ans_l = (ans >> (ql - 1)).count();
      int ans_r = (ans >> (qr - len + 1)).count();
      printf("%d\n", ans_l - ans_r);
    }
  }
}
}  // namespace Primary
int main() {
  Primary::main();
  return 0;
}