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;
const int N = 5e5+5;
int c[26][N];char s[N];
inline int lowbit(int x){return x&(-x);}
void add(int i,int x,int num){
    while(x<=N){
        c[i][x]+=num;
        x+=lowbit(x);
    }
}
int sum(int i,int x){
    int ans=0;
    while(x){
        ans+=c[i][x];
        x-=lowbit(x);
    }
    return ans;
}
int main(){
	int n,m,op,x,y;char y1[2];
	scanf("%d%s",&n,s+1);
	for(int i=1;s[i];i++)add(s[i]-'a',i,1);
	scanf("%d",&m);
	for(int i=0;i<m;i++){
		scanf("%d",&op);
		if(op==1){
			scanf("%d%s",&x,y1);
			add(s[x]-'a',x,-1);
			s[x]=*y1;add(s[x]-'a',x,1);
		} 
		else{
			int ans=0;
			scanf("%d%d",&x,&y);
			for(int i=0;i<26;i++)if(sum(i,y)-sum(i,x-1))ans++;
			printf("%d\n",ans);
		}
	}
	return 0;
}