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;
bitset<100001> bs[26], ba, bm;
char s[100001], t[100001];
int main(void) {
  scanf("%s", s);
  int n = strlen(s);
  for (int i = 0; i != n; ++i) bs[s[i] - 'a'][i] = bm[i] = 1;
  int q;
  scanf("%d", &q);
  while (q--) {
    int o;
    scanf("%d", &o);
    if (o == 1) {
      int i;
      char b[10];
      scanf("%d%s", &i, b);
      i--;
      bs[s[i] - 'a'][i] = 0;
      s[i] = b[0];
      bs[s[i] - 'a'][i] = 1;
    } else {
      int l, r, m;
      scanf("%d%d%s", &l, &r, t);
      m = strlen(t);
      ba.set();
      for (int i = 0; i != m; ++i) {
        ba &= (bs[t[i] - 'a'] >> i);
      }
      ba >>= l - 1;
      ba <<= (n - (r - l - m + 2));
      ba &= bm;
      printf("%d\n", ba.count());
    }
  }
  return 0;
}