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;
long long inf = 1000 * 1000 * 1000 + 7;
long long inf64 = inf * inf;
long long mod = 228228227;
long long math_mod = 998244353;
float START_TIME;
mt19937 gen(time(0));
bitset<100005> a[27], one, ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cerr << fixed << setprecision(15);
  cout << fixed << setprecision(15);
  srand(time(0));
  START_TIME = clock();
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); ++i) one[i] = 1;
  for (int i = 0; i < s.size(); ++i) a[s[i] - 'a'][i] = 1;
  int m;
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int k;
    cin >> k;
    if (k == 1) {
      int pos;
      char c;
      cin >> pos >> c;
      a[s[pos - 1] - 'a'][pos - 1] = 0;
      a[c - 'a'][pos - 1] = 1;
      s[pos - 1] = c;
    } else {
      int l, r;
      cin >> l >> r;
      string st;
      cin >> st;
      int len = st.size();
      if (len > r - l + 1) {
        cout << 0 << endl;
        continue;
      } else {
        ans = one;
        for (int i = 0; i < len; ++i) ans &= (a[st[i] - 'a'] >> i);
        int res = (ans >> (l - 1)).count() - (ans >> (r - len + 1)).count();
        cout << res << endl;
      }
    }
  }
  cerr << endl << (clock() - START_TIME) / CLOCKS_PER_SEC << " sec." << endl;
}