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
// E - Simple String Queries
// ordered set version
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)

int main(){
	int N,Q; string S; cin>>N>>S>>Q;
	vector<set<int>> abc(26);
	rep(i,N) abc[S[i]-'a'].insert(i);
	rep(i,Q){
		int t; cin>>t;
		if(t-1){//type2 display
			int l,r;cin>>l>>r; --l; --r;
			int ans = 0;
			rep(i,26){
				auto it = abc[i].lower_bound(l);
				if(it != abc[i].end() && *it <= r) ans++;
			}
			printf("%d\n",ans);
		}
		else{//type1 update
			int i; char c; cin>>i>>c; --i;
			if(S[i] == c) continue;
			abc[S[i]-'a'].erase(i);
			S[i] = c;
			abc[S[i]-'a'].insert(i);
		}
	}
}