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 ;

const int MAX = 27 ;

int n , q ;

string s ;

set<int>ss[MAX] ;

int main()
{
	ios_base::sync_with_stdio(0) ;
	cin.tie(0) ;
	cin>>n ;
	cin>>s ;
	cin>>q ;
	for(int i = 0 ; i < n ; ++i)
		ss[(s[i] - 'a')].insert(i) ;
	while(q--)
	{
		int t , x ; 
		cin>>t>>x ;
		--x ;
		if(t == 1)
		{
			char c ;
			cin>>c ;
			ss[(s[x] - 'a')].erase(x) ;
			s[x] = c ;
			ss[(s[x] - 'a')].insert(x) ;
		}
		else if(t == 2)
		{
			int y; 
			cin>>y ;
			--y ;
			int ans = 0 ;
			for(int i = 0 ; i < 26 ; ++i)
			{
				if(ss[i].size() == 0)
					continue ;
				int a = *ss[i].rbegin() ;
				if(a < x)
					continue ;
				a = *ss[i].lower_bound(x) ;
				if(a <= y)
					ans++ ;
			}
			cout<<ans<<"\n" ;
		}
	}
	return 0 ;
}		