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;
string s, r;
bitset<100010> inf;
bitset<100010> ans;
bitset<100010> a[26];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; ++i) {
    inf[i] = 1;
    a[s[i] - 'a'][i] = 1;
  }
  int q;
  cin >> q;
  r = s;
  while (q--) {
    int t;
    char c;
    cin >> t;
    if (t == 1) {
      int p;
      cin >> p >> c;
      --p;
      a[r[p] - 'a'][p] = 0;
      r[p] = c;
      a[c - 'a'][p] = 1;
      continue;
    }
    int l, r;
    cin >> l >> r >> s;
    if (s.size() > r - l + 1) {
      cout << 0 << '\n';
      continue;
    }
    --l;
    --r;
    ans = inf;
    for (int i = 0; i < s.size(); ++i) {
      ans &= (a[s[i] - 'a'] >> (l + i));
    }
    cout << ans.count() - (ans >> (r - l - s.size() + 2)).count() << '\n';
  }
  return 0;
}