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].
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - 48;
  return x * f;
}
inline void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar((x % 10) + '0');
}
inline void writeln(int x) {
  write(x);
  puts("");
}
const int oo = 0x3f3f3f3f;
const int inf = oo;
bitset<100005> f[26], ans;
char s[100005], ch[100005];
int n, Q;
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n; i++) s[i] -= 'a', f[s[i]][i] = 1;
  Q = read();
  while (Q--) {
    int op = read();
    if (op == 1) {
      int pos = read(), ch = getchar();
      f[s[pos]][pos] = 0;
      s[pos] = ch - 'a';
      f[s[pos]][pos] = 1;
    } else {
      int l = read(), r = read();
      scanf("%s", ch);
      int len = strlen(ch);
      if (r - l + 1 < len) {
        puts("0");
        continue;
      }
      ans.set();
      ans <<= (l - 1);
      ans ^= (ans << (r - l + 2 - len));
      for (int i = 0; i < len; i++) ans = (ans << 1) & f[ch[i] - 'a'];
      writeln(ans.count());
    }
  }
  return 0;
}
