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>
inline int sbt(int x) { return __builtin_popcount(x); }
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int n;
string s;
int q;
bitset<100013> bit[27];
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  cin >> s;
  s = '$' + s;
  for (int i = 1; i < s.size(); i++) {
    bit[s[i] - 'a'].set(i, 1);
  }
  const int nm = s.size() - 1;
  bitset<100013> o;
  for (int i = 0; i < nm; i++) o.set(i);
  cin >> q;
  while (q--) {
    int ty;
    cin >> ty;
    if (ty == 1) {
      int i;
      char c;
      cin >> i >> c;
      bit[s[i] - 'a'].set(i, 0);
      bit[c - 'a'].set(i, 1);
      s[i] = c;
    } else {
      bitset<100013> res = o;
      int l, r;
      string t;
      cin >> l >> r >> t;
      if ((r - l + 1) < t.size()) {
        cout << 0 << endl;
        continue;
      }
      t = '$' + t;
      int n = t.size();
      for (int i = 1; i < n; i++) {
        res = (res & (bit[t[i] - 'a'] >> i));
      }
      if ((r - (n - 1) + 1) < 0) {
        cout << 0 << endl;
        continue;
      }
      cout << ((res >> (l - 1)).count()) - ((res >> (r - (n - 1) + 1)).count())
           << endl;
    }
  }
  return 0;
}