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;
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, q, opt, x, y;
char s[100005], t[100005];
bitset<100005> b[30], now;
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n; i++) b[s[i] - 'a' + 1][i] = 1;
  q = readint();
  char ch;
  while (q--) {
    opt = readint();
    x = readint();
    if (opt == 1) {
      scanf("%c", &ch);
      b[s[x] - 'a' + 1][x] = 0, b[ch - 'a' + 1][x] = 1;
      s[x] = ch;
    } else {
      y = readint();
      scanf("%s", t + 1);
      m = strlen(t + 1);
      now = b[t[1] - 'a' + 1] << (100005 - y - 1);
      for (int i = 2; i <= m; i++)
        now &= (b[t[i] - 'a' + 1] << (100005 - y - 1) >> (i - 1));
      printf("%d\n", (now >> (100005 - y - 1 + x)).count());
    }
  }
  return 0;
}