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 N=5e5+5;
int n,Q;
char s[N];
class BIT{
public:
	int sum[N];
	void ins(const int &x,const int &v){
		for(int i=x;i<=n;i+=i&-i)sum[i]+=v;
	}
	int ask(const int &x)const {
		int ans=0;
		for(int i=x;i;i&=i-1)ans+=sum[i];
		return ans;
	}
}bit[26];
int main(){
	cin>>n;
	scanf("%s",s+1);
	for(int i=1;i<=n;i++)bit[s[i]-'a'].ins(i,1);
	scanf("%d",&Q);
	while(Q--){
		int op;
		scanf("%d",&op);
		if(op==1){
			int x;char c;
			scanf("%d %c",&x,&c);
			if(s[x]==c)continue;
			bit[s[x]-'a'].ins(x,-1);
			bit[c-'a'].ins(x,1);
			s[x]=c;
		}else{
			int l,r;
			scanf("%d%d",&l,&r);
			int ans=0;
			for(int i=0;i<26;i++)ans+=bit[i].ask(r)-bit[i].ask(l-1)>0;
			printf("%d\n",ans);
		}
	}		
}
