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>
#define rep(i,n) for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair <int,int> P;


int main(){
  int n,q;
  string s;
  cin>>n>>s>>q;
  set<int> se[26];
  rep(i,n){
    se[s[i]-'a'].insert(i);
  }
  rep(i,q){
    int t;
    cin>>t;
    if(t==1){
      int j;
      char c;
      cin>>j>>c;
      j--;
      se[s[j]-'a'].erase(j);
      se[c-'a'].insert(j);
      s[j]=c;
    }
    else{
      int l,r;
      cin>>l>>r; l--; r--;
      int ans=0;
      rep(j,26){
	if(se[j].size()>0&&se[j].lower_bound(l)!=se[j].end()
	   &&*se[j].lower_bound(l)<=r){
	  ans++;
	}
      }
      cout<<ans<<endl;
    }
  }
  return 0;
}



