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 <stdio.h>
#include <string.h>
#define maxn 500010
int c[30][maxn],n;
char s[maxn];
int lowbit(int x){
	return x&-x;
}
void update(int x,int delta,int *a){
	for(int i=x;i<=n;i+=lowbit(i))a[i]+=delta;
}
int sum(int x,int *a){
	int ret=0;//此处错写成int ret;查了好久
	for(int i=x;i>=1;i-=lowbit(i))ret+=a[i];
	return ret;
}
int main(){
	int i,q,cmd,p,l,r,cnt,j;
	char b[3];
	scanf("%d%s%d",&n,s+1,&q);
	for(i=1;i<=n;i++)update(i,1,c[s[i]-'a']);
	while(q--){
		scanf("%d",&cmd);
		if(cmd==1){
			scanf("%d%s",&p,b);
			update(p,-1,c[s[p]-'a']);
			s[p]=b[0],update(p,1,c[b[0]-'a']);
		}else if(cmd==2){
			scanf("%d%d",&l,&r),cnt=0;
			for(i=0;i<26;i++)cnt+=sum(r,c[i])-sum(l-1,c[i])>0;
			printf("%d\n",cnt);
		}
	}
}