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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#define rd(X) X=0;for(;;){char k=buf[sz++];if(k<'0')break;X=X*10+k-'0';}
char buf[1<<20];sz;
si;N,Q;
char out[2<<17];outsz;
ap[128];
main()
{
	read(0,buf,1<<20);
	rd(N)
	si=sz-1;
	while(buf[sz++]!='\n');
	rd(Q)
	for(int i=0;i++<Q;)
	{
		if(buf[sz]=='1')
		{
			sz+=2;
			int id;
			rd(id)
			buf[si+id]=buf[sz++];
			sz++;
		}
		else
		{
			sz+=2;
			int l,r;
			rd(l)
			l+=si;
			rd(r)
			r+=si;
			int ans=0;
			for(int j=l;j<=r;j++)
			{
				if(ap[buf[j]]<i)
				{
					ap[buf[j]]=i;
					ans++;
					if(ans==26)break;
				}
			}
			if(ans<10)
			{
				out[outsz++]=ans+'0';
			}
			else
			{
				out[outsz++]=ans/10+'0';
				out[outsz++]=ans%10+'0';
			}
			out[outsz++]='\n';
		}
	}
	write(1,out,outsz);
}