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;
inline int getInt() {
  int res(0);
  char c = getchar();
  while (c < '0') c = getchar();
  while (c >= '0') {
    res = res * 10 + (c - '0');
    c = getchar();
  }
  return res;
}
inline int fastpo(int x, int n, int mod) {
  int res(1);
  while (n) {
    if (n & 1) {
      res = res * (long long)x % mod;
    }
    x = x * (long long)x % mod;
    n /= 2;
  }
  return res;
}
const int N = 100000;
const int LOG = 20;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
bitset<N> msk[26];
int main() {
  string st;
  cin >> st;
  for (int i(0); i < (int)st.size(); i++) {
    msk[st[i] - 'a'].set(i);
  }
  int Q;
  scanf("%d", &Q);
  for (int i(0); i < Q; i++) {
    int tp;
    scanf("%d", &tp);
    if (tp == 2) {
      int le, ri;
      scanf("%d%d", &le, &ri);
      le--;
      ri--;
      string a;
      cin >> a;
      if (ri - le + 1 < (int)a.size()) {
        printf("0\n");
        continue;
      }
      bitset<N> tmp;
      tmp.set();
      for (int i(0); i < (int)a.size(); i++)
        tmp = tmp & (msk[a[i] - 'a'] >> (le + i));
      printf("%d\n",
             (int)(tmp << (N - (ri - le + 1 - (int)a.size() + 1))).count());
    } else {
      int le;
      char c;
      scanf("%d %c", &le, &c);
      msk[st[le - 1] - 'a'].set(le - 1, 0);
      msk[c - 'a'].set(le - 1, 1);
      st[le - 1] = c;
    }
  }
}