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;
bitset<100005> mask[26];
void test() {
  for (int i = 0; i < 26; i++) {
    cout << (char)(i + 'a') << " " << mask[i] << endl;
  }
}
void computeMask(string &text) {
  for (int i = 0; i < (int)text.size(); ++i) {
    int c = text[i] - 'a';
    mask[c].set(i);
  }
}
int StringMatchingInRange(string &pattern, string &text, long long l,
                          long long r) {
  if ((int)pattern.size() > (r - l + 1)) return 0;
  bitset<100005> startMask;
  startMask.set();
  for (int i = 0; i < (int)pattern.size(); ++i) {
    int c = pattern[i] - 'a';
    startMask &= (mask[c] >> i);
  }
  return (startMask >> (l)).count() -
         (startMask >> (r - (int)pattern.size() + 2)).count();
}
int StringMatching(string &pattern, string &text) {
  if ((int)pattern.size() > (int)text.size()) return 0;
  bitset<100005> startMask;
  startMask.set();
  for (int i = 1; i < (int)pattern.size(); ++i) {
    int c = pattern[i] - 'a';
    startMask &= (mask[c] >> i);
  }
  return startMask.count();
}
void update(int idx, char ch, string &text) {
  char old = text[idx];
  mask[old - 'a'][idx] = 0;
  text[idx] = ch;
  mask[ch - 'a'][idx] = 1;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s, t;
  cin >> t;
  computeMask(t);
  int q, ty, l, r, x;
  char ch;
  cin >> q;
  while (q--) {
    cin >> ty;
    if (ty == 2) {
      cin >> l >> r >> s;
      l--, r--;
      cout << StringMatchingInRange(s, t, l, r) << endl;
    } else {
      cin >> x >> ch;
      x--;
      update(x, ch, t);
    }
  }
  return 0;
}