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;
template <class t>
inline t read(t &x) {
  x = 0;
  char c = getchar();
  bool f = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t>
inline void write(t x) {
  if (x < 0) {
    putchar('-'), write(-x);
  } else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const int N = 1e5 + 5;
bitset<N> a[30], ans;
char s[N], ss[N];
int q;
signed main() {
  scanf("%s", s);
  for (int i = 0; s[i]; i++) a[s[i] - 'a'][i] = 1;
  read(q);
  while (q--) {
    int op;
    read(op);
    if (op == 1) {
      int id;
      char c;
      read(id);
      scanf(" %c", &c);
      id--;
      a[s[id] - 'a'][id] = 0;
      a[c - 'a'][id] = 1;
      s[id] = c;
    } else {
      int l, r;
      read(l);
      l--;
      read(r);
      r--;
      scanf("%s", ss);
      ans.set();
      for (int i = 0; ss[i]; i++) ans &= a[ss[i] - 'a'] >> i;
      int rn = (ans >> l).count(),
          ln = (ans >> (r - (strlen(ss) - 1) + 1)).count();
      write(max(0, rn - ln));
      puts("");
    }
  }
}