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;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << '\n';
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
void run() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
}
const int N = 1e5 + 5, mod = 1e9 + 7, mod2 = 998244353, MAXN = 100000 + 29;
void files() {}
bitset<N> bits[26];
int countt(bitset<N> &a, int l) {
  int shift = N - l;
  return (a << shift).count();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  files();
  string s;
  cin >> s;
  int q;
  cin >> q;
  for (int i = 0; i < ((int)((s).size())); ++i) {
    bits[s[i] - 'a'].set(i);
  }
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int i;
      cin >> i;
      i--;
      char c;
      cin >> c;
      bits[s[i] - 'a'].reset(i);
      s[i] = c;
      bits[s[i] - 'a'].set(i);
    } else {
      int l, r;
      cin >> l >> r;
      string sub;
      cin >> sub;
      if (r - l + 1 < ((int)((sub).size()))) {
        cout << 0 << '\n';
        continue;
      }
      bitset<N> ans;
      ans.set();
      for (int i = 0; i < ((int)((sub).size())); ++i) {
        ans &= (bits[sub[i] - 'a'] >> i);
      }
      int shift = N - r + ((int)((sub).size())) - 1;
      int res = (ans << shift).count();
      shift = N - l + 1;
      res -= (ans << shift).count();
      cout << res << '\n';
    }
  }
}