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;

int ft[27][500005];
int n,q;
char s[500005];
char sb[10];

void upd( int c, int x, int v ){
	while ( x <= n ){
		ft[c][x] += v;
		x += x & (-x);
	}
}

int rsq( int c, int x) {
	int ans = 0;
	while ( x > 0 ){
		ans += ft[c][x];
		x -= x & (-x);
	}
	return ans;
}

int main(){
	
	scanf("%d%s%d",&n,&s,&q);
	for ( int i = 0; i < n; i++ ){
		int v = s[i] - 'a';
		upd(v, i + 1, 1);
	}
	for ( int i = 0; i < q; i++ ){
		int c, a,b;
		scanf("%d%d",&c,&a);
		if ( c == 1 ){
			scanf("%s",&sb);
			int v = s[a-1] -'a';
			upd(v, a  , - 1 );
			s[a-1] = sb[0];
			upd(sb[0]-'a', a , 1 );
		} else {
			scanf("%d",&b);
			int ans = 0;
			for ( int i = 0; i < 26; i++ ){
				if ( rsq(i, b) - rsq(i,a-1) > 0 ) ans++;
			}
			printf("%d\n",ans);
		}
	}
	return 0;
}