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 comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long N = 100001;
const long long MOD = 1000000007;
const long long INF = LLONG_MAX;
bitset<N> bs[26], all, ans;
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL);
  int t = 1, cs = 0;
  while (t--) {
    int n, q, type, id, i, l, r;
    string s, p;
    char ch;
    cin >> s >> q;
    n = s.size();
    for (i = 0; i < n; i++) {
      all[i] = 1;
      bs[s[i] - 'a'][i] = 1;
    }
    while (q--) {
      cin >> type;
      if (type == 1) {
        cin >> id >> ch;
        id--;
        bs[s[id] - 'a'][id] = 0;
        bs[ch - 'a'][id] = 1;
        s[id] = ch;
      } else {
        cin >> l >> r >> p;
        if (p.size() > r - l + 1) {
          cout << 0 << endl;
          continue;
        }
        ans = all;
        for (i = 0; i < p.size(); i++) ans &= (bs[p[i] - 'a'] >> i);
        cout << (ans >> (l - 1)).count() - (ans >> (r - p.size() + 1)).count()
             << endl;
      }
    }
  }
}