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;
int Bit[26][N+100];
char str[N+100];

void update(int id,int idx,int val)
{
	while(idx<=N)
	{
		Bit[id][idx] += val;
		idx += (idx&-idx);
	}
}
int query(int id,int idx)
{
	int total = 0;
	while(idx>0)
	{
		total += Bit[id][idx];
		idx -= (idx&-idx);
	}
	return total;
}
int main()
{
	int n,q;
	scanf("%d%s%d",&n,str+1,&q);
	for(int i=1;i<=n;i++){
		update(str[i]-'a',i,1);
	}
	while(q--)
	{
		int type;
		scanf("%d",&type);
		if(type==1){
			int idx;
			char s[3];
			scanf("%d%s",&idx,s);
			update(str[idx]-'a',idx,-1);
			update(s[0]-'a',idx,1);
			str[idx] = s[0];
		}
		else{
			int l,r,ans = 0;
			scanf("%d%d",&l,&r);
			for(int i=0;i<26;i++){
				if(query(i,r)-query(i,l-1)){
					ans++;
				}
			}
			printf("%d\n",ans);
		}
	}
}
