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>
#define pb push_back
#define mp make_pair
#define int long long
#define inf 1000000007
#define LINF 100000000000000007LL
#define ll long long
using namespace std;
signed main(){
	int n,q;
	string st;
	cin>>n>>st>>q;
	set<int> idx[30];
	for(int i=1;i<=n;i++){
		idx[ st[i-1] - 'a' ].insert(i);
	}
	for(int i=0;i<q;i++){
		int tp;
		cin>>tp;
		if( tp == 1 ){
			int pla;
			char ch;
			cin>>pla>>ch;
			idx[ st[pla-1] - 'a' ].erase(pla);
			st[pla-1] = ch;
			idx[ st[pla-1] - 'a' ].insert(pla);
		}
		else{
			int lef,rig;
			cin>>lef>>rig;
			int cnt = 0;
			for(int i=0;i<26;i++){
				auto it = idx[i].lower_bound(lef);
				if( it != idx[i].end() && *it <= rig ) cnt ++ ;
			}
			cout<<cnt<<endl;
		}
	}
	return 0;
}