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 ch = getchar();
  bool f = 0;
  int x = 0;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = 1;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ 48);
  if (f == 1) x = -x;
  return x;
}
int len, num, opt, l, r, x, n, ans1, ans2, len1;
bitset<100005> a[27], b;
char s[100005], c;
int main() {
  scanf("%s", s);
  len1 = strlen(s);
  for (int i = 0; i < len1; i++) {
    a[s[i] - 'a'][i + 1] = 1;
  }
  n = read();
  for (int i = 1; i <= n; i++) {
    opt = read();
    if (opt == 1) {
      scanf("%d %c", &x, &c);
      a[s[x - 1] - 'a'][x] = 0;
      a[c - 'a'][x] = 1;
      s[x - 1] = c;
    } else {
      l = read(), r = read();
      char s2[100005] = {0};
      scanf("%s", s2);
      len = strlen(s2);
      b.set();
      for (int j = 0; j < len; j++) {
        b = b & (a[s2[j] - 'a'] >> j);
      }
      ans1 = (b >> l).count(), ans2 = (b >> (r - len + 2)).count();
      if (ans1 - ans2 < 0) ans1 = ans2;
      printf("%d\n", ans1 - ans2);
    }
  }
  return 0;
}