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;
typedef long long ll;
const ll maxn = 1e6 + 5;

int n,Q;
char s[maxn];
set<int>ss[30];

int main(){
	cin>>n>>s+1>>Q;
	for(int i=0;i<26;i++) 
        ss[i].insert(n+1);
	for(int i=1;i<=n;i++) 
        ss[s[i]-'a'].insert(i);
	
    while(Q--){
		int t,i,l,r;
		char c;
		cin>>t;
		if(t==1){
			cin>>i>>c;
			if(s[i]!=c) 
                ss[s[i]-'a'].erase(i);
			s[i]=c;
			ss[c-'a'].insert(i);
		}else{
			int ans=0;
			cin>>l>>r;
			for(int i=0;i<26;i++)
				if(*ss[i].lower_bound(l)<=r) 
                    ans++;
			cout<<ans<<'\n';
		}
	}
	return 0;
}