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<iostream>
#include<cstdio>
#include<cstring>
#include<set>
using namespace std;
const int maxn=500009;
set<int> ss[26];
int main()
{
	char s[maxn];
	int n,q,iq,lq,rq,cq;
	cin>>n;
	getchar();
	for(int i=1;i<=n;i++)
	{
		scanf("%c",&s[i]);
		ss[s[i]-'a'].insert(i);
	}
	cin>>q;
	int f;
	while(q--)
	{
		scanf("%d",&f);
		if(f==1)
		{
		scanf("%d",&iq);
		getchar();
		scanf("%c",&cq);
		ss[s[iq]-'a'].erase(iq);
		s[iq]=cq;//
		ss[cq-'a'].insert(iq);
	    }
	    else
	    {
	    	scanf("%d%d",&lq,&rq);
	    	int num=0;
	    	set<int>::iterator it;
			for(int i=0;i<26;i++)
			{
				it=ss[i].lower_bound(lq);
				if(it!=ss[i].end()&&*it<=rq)
				num++;
			}
			printf("%d\n",num);
		}
	}
	return 0;
}