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 main() {
	int n,q,i,j,query,c,l,r,ans;
	string s;
	char ch1,ch2;
	vector<set<int>> a(26);
	cin >> n >> s >> query;
	for(i=1; i<=n; i++){
		c=s[i-1]-'a';
		a[c].insert(i);
	}
	
	for(c=0; c<26; c++){
		a[c].insert(500001);
	}
	
	for(q=0; q<query; q++){
		cin >> l;
		if(l==1){
			cin >> r >> ch2;
			c=s[r-1]-'a';
			a[c].erase(r);
			s[r-1]=ch2;
			c=ch2-'a';
			a[c].insert(r);
		}else{
			cin >> l >> r;
			ans=0;
			for(c=0; c<26; c++){
				auto itr=a[c].begin();
				itr=a[c].lower_bound(l);
				if(*itr<=r){
					ans++;
				}
			}
			cout << ans << endl;
		}
		
	}
	return 0;
}