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
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)

int N,Q;
string S;
vector<set<int>> abc(26);

void add(int i, char c){//type1
	if(S[i] == c) return;
	abc[S[i]-'a'].erase(i);
	S[i] = c;
	abc[S[i]-'a'].insert(i);
}

void out(int l, int r){//type2
	int ans = 0;
	rep(i,26){
		if(abc[i].size() <= 0) continue;
		auto it = abc[i].lower_bound(l);
		if(it != abc[i].end() && *it <= r) ans++;
	}
	printf("%d\n",ans);
}

int main(){
	cin>>N>>S>>Q;
	rep(i,N) abc[S[i]-'a'].insert(i);
	rep(i,Q){
		int t,a;
		cin>>t;
		if(t-1){//type2
			int b;
			cin>>a>>b;
			out(a-1, b-1);
		}
		else{//type1
			char b;
			cin>>a>>b;
			add(a-1, b);
		}
	}
}