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>
#pragma GCC optimize(2)
using namespace std;
int read() {
  char ch = getchar();
  int w = 1, c = 0;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') w = -1;
  for (; isdigit(ch); ch = getchar()) c = (c << 3) + (c << 1) + (ch ^ 48);
  return w * c;
}
const int M = 1e5 + 10;
bitset<M> s[30], ans;
char ch[M], t[M];
int n, m;
int main() {
  ios::sync_with_stdio(0);
  cin >> (ch + 1);
  n = strlen(ch + 1);
  for (int i = 1; i <= n; ++i) {
    s[ch[i] - 'a'][i - 1] = 1;
  }
  cin >> m;
  while (m--) {
    int ty, x, y, z;
    cin >> ty >> x;
    if (ty == 1) {
      char c;
      cin >> c;
      s[ch[x] - 'a'][x - 1] = 0;
      s[c - 'a'][x - 1] = 1;
      ch[x] = c;
    } else {
      cin >> y;
      cin >> (t + 1);
      int l = strlen(t + 1);
      ans.set();
      for (int i = 1; i <= l; ++i) {
        ans &= (s[t[i] - 'a'] >> (i - 1));
      }
      int lans = (ans >> (x - 1)).count(), rans = (ans >> (y - l + 1)).count();
      cout << max(lans - rans, 0) << "\n";
    }
  }
  return 0;
}