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<algorithm>
#include<set>
 using namespace std;
 set<int> a[26];
 string temp;
 int main()
 {
 	int n,op,pos,l,r,m;
 	cin>>n;
	cin>>temp;
	char ch;
	for(int i=0;i<temp.size();i++) a[temp[i]-'a'].insert(i+1);
	scanf("%d",&m);
	while(m--){
		scanf("%d",&op);
		if(op==1){
			cin>>pos>>ch;
			a[temp[pos-1]-'a'].erase(pos);
			temp[pos-1]=ch;
			a[temp[pos-1]-'a'].insert(pos);
		}
		else{
			cin>>l>>r;
			int num=0;
			for(int i=0;i<26;i++){
				set<int>::iterator it;
				it=a[i].lower_bound(l);
				if(it!=a[i].end()&&*it<=r) num++;
			}
			cout<<num<<endl;
		}
	}
	return 0;
 }