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;
namespace {
const long long MAX = 100002;
bitset<MAX> was[26];
void Solve(long long test) {
  (void)test;
  string str;
  cin >> str;
  for (long long i = (long long)(0); i < (long long)(((long long)(str.size())));
       ++i) {
    was[str[i] - 'a'].set(i);
  }
  long long q;
  cin >> q;
  for (long long _ = (long long)(0); _ < (long long)(q); ++_) {
    long long ty;
    cin >> ty;
    if (ty == 1) {
      long long i;
      char c;
      cin >> i >> c;
      --i;
      was[str[i] - 'a'].flip(i);
      str[i] = c;
      was[str[i] - 'a'].flip(i);
    } else {
      long long l, r;
      string y;
      cin >> l >> r >> y;
      --l, --r;
      if (((long long)(y.size())) > r - l + 1) {
        cout << 0 << '\n';
        continue;
      }
      bitset<MAX> mask, ones;
      mask.flip();
      ones.flip();
      ones <<= (r - l - ((long long)(y.size())) + 2);
      ones.flip();
      for (long long i = (long long)(0);
           i < (long long)(((long long)(y.size()))); ++i) {
        mask &= (was[y[i] - 'a'] >> i);
      }
      mask >>= l;
      mask &= ones;
      const long long res = mask.count();
      cout << res << '\n';
    }
  }
}
void Init() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
}
}  // namespace
int main() {
  Init();
  long long tests = 1;
  for (long long test = (long long)(1); test < (long long)(tests + 1); ++test)
    Solve(test);
  return 0;
}