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 read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
inline char readchar() {
  char c = getchar();
  while (c < 'a' || c > 'z') c = getchar();
  return c;
}
const int N = 100005;
int n;
char str[N], nowstr[N];
bitset<N> v[26], ans;
int main() {
  scanf("%s", str + 1);
  n = strlen(str + 1);
  for (int i = 1; i <= n; i++) v[str[i] - 'a'].set(i, 1);
  int T = read();
  while (T--) {
    int opt = read();
    if (opt == 1) {
      int x = read();
      char nc = readchar();
      if (str[x] != nc)
        v[nc - 'a'].set(x, 1), v[str[x] - 'a'].set(x, 0), str[x] = nc;
    } else {
      int l = read(), r = read();
      scanf("%s", nowstr + 1);
      int sl = strlen(nowstr + 1);
      ans.set();
      for (int i = 1; i <= sl; i++) ans &= (v[nowstr[i] - 'a'] >> (i - 1));
      int lc = (ans >> l).count(), rc = (ans >> (r - sl + 2)).count();
      if (lc < rc)
        printf("0\n");
      else
        printf("%d\n", lc - rc);
    }
  }
  return 0;
}