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;
inline int getint() {
  int num = 0, flag = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag = -1;
  while (c >= '0' && c <= '9') num = num * 10 + c - 48, c = getchar();
  return num * flag;
}
int n, m;
char s[100005], t[100005];
bitset<100005> a[27], ans;
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; i++) a[s[i] - 97][i + 1] = 1;
  m = getint();
  while (m--) {
    int op = getint();
    if (op == 1) {
      int k = getint() - 1;
      scanf("%s", t);
      a[s[k] - 97][k + 1] = 0, a[t[0] - 97][k + 1] = 1, s[k] = t[0];
    } else {
      int l = getint(), r = getint();
      scanf("%s", t), n = strlen(t);
      ans.set();
      for (int i = 0; i < n; i++) ans &= (a[t[i] - 97] >> i);
      int L = (ans >> l).count(), R = (ans >> (r - n + 2)).count();
      printf("%d\n", max(L - R, 0));
    }
  }
}