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<100010> a[26], ans;
int cnt;
vector<int> ori;
void cg() {
  int pos;
  char targ;
  scanf("%d %c", &pos, &targ);
  a[ori[pos - 1]][pos] = 0;
  ori[pos - 1] = targ - 'a';
  a[ori[pos - 1]][pos] = 1;
}
char ta[100010];
void qr() {
  int l, r;
  scanf("%d%d%s", &l, &r, ta);
  int m = strlen(ta);
  if (r - l + 1 < m) {
    printf("0\n");
    return;
  }
  ans.set();
  for (int i = 0; i < m; i++) {
    ans &= (a[ta[i] - 'a'] >> i);
  }
  int res = (ans >> (l)).count() - (ans >> (r - m + 2)).count();
  printf("%d\n", res);
}
int main() {
  int q;
  char tmp;
  while ((tmp = getchar()) != '\n') {
    int ch = tmp - 'a';
    ori.push_back(ch);
    cnt++;
    a[ch][cnt] = 1;
  }
  scanf("%d", &q);
  while (q--) {
    int kind;
    scanf("%d", &kind);
    if (kind == 1)
      cg();
    else
      qr();
  }
  return 0;
}