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(){
  int N;
  cin >> N;
  string S;
  cin >> S;
  vector<set<int>> vec(26);
  for(int i=0; i<N; i++){
    int x = S.at(i) - 'a';
    vec.at(x).insert(i);
  }
  int Q;
  cin >> Q;
  for(int i=0; i<Q; i++){
    int a;
    cin >> a;
    if(a == 1){
      int b;
      cin >> b;
      b--;
      char c;
      cin >> c;
      int x = S.at(b) - 'a';
      int y = c - 'a';
      S.at(b) = c;
      vec.at(x).erase(b);
      vec.at(y).insert(b);
    }
    else{
      int cnt = 0;
      int b,c;
      cin >> b >> c;
      b--; c--;
      for(int j=0; j<26; j++){
        if(vec.at(j).set::lower_bound(b) != vec.at(j).set::lower_bound(c+1)){
          cnt++;
        }
      }
      cout << cnt << endl;
    }
  }
} 