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;
const long long inf = 2e9;
const long long mod = 998244353;
const int maxn = 1e5 + 50;
bitset<maxn> s[27], ans;
char a[maxn], b[maxn];
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> a + 1;
  int len = strlen(a + 1);
  for (int i = 1; i <= len; i++) s[a[i] - 'a'].set(i);
  int q;
  cin >> q;
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int x;
      char ch;
      cin >> x >> ch;
      s[a[x] - 'a'][x] = 0;
      s[(a[x] = ch) - 'a'][x] = 1;
    } else {
      ans.set();
      int l, r;
      cin >> l >> r;
      cin >> b;
      int len1 = strlen(b);
      if (len1 > (r - l + 1)) {
        cout << 0 << '\n';
        continue;
      }
      for (int i = 0; i < len1; i++) {
        ans &= (s[b[i] - 'a'] >> i);
      }
      int num1 = (ans >> (l)).count();
      int num2 = (ans >> (r - len1 + 2)).count();
      cout << num1 - num2 << '\n';
    }
  }
  return 0;
}