You are given a string S of length N consisting of lowercase English letters.

Process Q queries of the following two types:

* Type 1: change the i_q-th character of S to c_q. (Do nothing if the i_q-th character is already c_q.)
* Type 2: answer the number of different characters occurring in the substring of S between the l_q-th and r_q-th characters (inclusive).

Constraints

* N, Q, i_q, l_q, and r_q are integers.
* S is a string consisting of lowercase English letters.
* c_q is a lowercase English letter.
* 1 \leq N \leq 500000
* 1 \leq Q \leq 20000
* |S| = N
* 1 \leq i_q \leq N
* 1 \leq l_q \leq r_q \leq N
* There is at least one query of type 2 in each testcase.

Input

Input is given from Standard Input in the following format:


N
S
Q
Query_1
\vdots
Query_Q


Here, Query_i in the 4-th through (Q+3)-th lines is one of the following:


1 i_q c_q



2 l_q r_q


Output

For each query of type 2, print a line containing the answer.

Example

Input

7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7


Output

3
1
5
#include <bits/stdc++.h>
using namespace std;

int main(){
  const int Nab= 26;
  int N;
  string str;
  cin >> N >> str;

  vector<set<int>> data(Nab);
  for(int i=0; i<N; i++){
    data.at(str.at(i)-'a').insert(i+1); // 1-indexed
  }

  int Q;
  cin >> Q;

  int query,buf1,buf2;
  char bufc;
  for(int i=0; i<Q; i++){
    cin >> query;
    if(query==1){
      cin >> buf1 >> bufc;
      for(int j=0; j<Nab; j++){
        data.at(j).erase(buf1);
      }
      data.at(bufc-'a').insert(buf1);
    }
    else{
      cin >> buf1 >> buf2;
      int ans= 0;
      for(int j=0; j<Nab; j++){
        auto itr1= data.at(j).lower_bound(buf1);
        auto itr2= data.at(j).upper_bound(buf2);
        if(distance(itr1,itr2) > 0){ ans++; }
      }
      cout << ans << endl;
    }
  }
}
