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 n,q,c[26][500009];
string s;
void update(int x,int v,int i) {
	for(;x<=n;x+=x&-x) c[i][x]+=v;
}
int sum(int x,int i) {
	int ans=0;
	for(;x>0;x-=x&-x) ans+=c[i][x];
	return ans;
}
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n>>s>>q;
	for(int i=0;i<s.size();i++) update(i+1,1,s[i]-'a');
	for(int i=0;i<q;i++) {
		int type;
		cin>>type;
		if(type==1) {
			int a;
			char b;
			cin>>a>>b;
			update(a,-1,s[a-1]-'a');
			s[a-1]=b;
			update(a,1,b-'a');
		}
		else {
			int l,r,ans=0;
			cin>>l>>r;
			for(int j=0;j<26;j++)
				if(sum(r,j)-sum(l-1,j)>0)
					ans++;
			cout<<ans<<endl;
		}
	}
	return 0;
}