Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7

Sample Output 1

3
1
5

In the first query, cdbb contains three kinds of letters: b , c , and d, so we print 3.

In the second query, S is modified to abcdzbd.

In the third query, a contains one kind of letter: a, so we print 1.

In the fourth query, S is modified to abcazbd.

In the fifth query, S does not change and is still abcazbd.

In the sixth query, abcazbd contains five kinds of letters: a, b, c, d, and z, so we print 5.

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
int n,m,a[100005],b[100005],sum[100005];
string s;
void build(int x,int l,int r){
	if(l==r){
		b[x]=a[l],sum[x]=1<<b[x];
		return ;
	}
	int mid=l+r>>1;
	build(x*2,l,mid);
	build(x*2+1,mid+1,r);
	sum[x]=sum[x*2]|sum[x*2+1];
}
void gao(int x,int l,int r,int l1,int w){
	if(l==l1 and r==l1){
		b[x]=w,sum[x]=1<<w;
		return ;
	}
	int mid=(l+r)>>1;
	if(l1<=mid)gao(x*2,l,mid,l1,w);
	else gao(x*2+1,mid+1,r,l1,w);
	sum[x]=sum[x*2]|sum[x*2+1];
}
int query(int x,int l,int r,int l1,int r1){
	if(l>=l1 and r<=r1)return sum[x];
	int res=0,mid=(l+r)>>1;
	if(l1<=mid)res|=query(x<<1,l,mid,l1,r1);
	if(r1>mid)res|=query(x<<1|1,mid+1,r,l1,r1);
	return res;
}
int main(){
	n=read();
	cin >> s;
	for(int i=0;i<s.size();i++)
		a[i+1]=s[i]-'a'+1;
	build(1,1,n);
	m=read();
	while(m--){
		int ab=read(),x=read(),y;
		if(ab==1){
			y=getchar();
			while(y==' ')y=getchar();
			y-='a'-1;
			gao(1,1,n,x,y);
		}
		else{
			y=read();
			printf("%d\n",__builtin_popcount(query(1,1,n,x,y)));
		}
	}
	return 0;
}