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;
const int maxn = 100010, inf = 1e9;
int n, ty, l, r, x;
bitset<maxn> v[26], ans;
char s[maxn], s2[maxn], c;
inline void read(int &k) {
  int f = 1;
  k = 0;
  char c = getchar();
  while (c < '0' || c > '9') c == '-' && (f = -1), c = getchar();
  while (c <= '9' && c >= '0') k = k * 10 + c - '0', c = getchar();
  k *= f;
}
int main() {
  scanf("%s", s + 1);
  int len = strlen(s + 1);
  for (int i = 1; i <= len; i++) v[s[i] - 'a'][i] = 1;
  read(n);
  for (int i = 1; i <= n; i++) {
    read(ty);
    if (ty == 1) {
      read(x);
      scanf("%c", &c);
      v[s[x] - 'a'][x] = 0;
      v[(s[x] = c) - 'a'][x] = 1;
    } else {
      read(l);
      read(r);
      scanf("%s", s2);
      int m = strlen(s2);
      if (r - l + 1 < m) {
        puts("0");
        continue;
      }
      ans.set();
      for (int j = 0; j < m; j++) ans &= (v[s2[j] - 'a'] >> j);
      printf("%d\n",
             (int)(ans >> (l)).count() - (int)(ans >> (r - m + 2)).count());
    }
  }
}