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 <stdio.h>
#include <utility>
#include <algorithm>
#include <set>
using namespace std;
char txt[500001];
set<int> visit[26];
int main(){
	int N;
	scanf("%d",&N);
	scanf("%s",txt+1);
	for(int i=1;i<=N;i++) visit[txt[i]-'a'].insert(i);
	for(int i=0;i<26;i++) visit[i].insert(987654321);
	int Q;scanf("%d",&Q);
	while(Q--){
		int a;scanf("%d",&a);
		if(a==1){
			int i;
			char c;
			scanf("%d %c",&i,&c);
			visit[txt[i]-'a'].erase(i);
			txt[i]=c;
			visit[txt[i]-'a'].insert(i);
		}
		else{
			int l,r;
			scanf("%d %d",&l,&r);
			int cnt=0;
			for(int i=0;i<26;i++){
				if((*visit[i].lower_bound(l))<=r) cnt++;
			}
			printf("%d\n",cnt);
		}
	}
}