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 mn=5e5+7;
int a[mn][26],n;
char s[mn];

int lowbit(int x) {return x&-x;}

void modify(int dl,int x,int d)
{
	while(x<=n) {
		a[x][d]+=dl;
		x+=lowbit(x);
	}
}

int query(int x,int d)
{
	int ans=0;
	while(x) {
		ans+=a[x][d];
		x-=lowbit(x);
	}
	return ans;
}

int main()
{
	int q,op,l,r,ans;
	scanf("%d%s%d",&n,s+1,&q);
	char c;
	for(int i=1;i<=n;++i) modify(1,i,s[i]-'a');
	for(int i=1;i<=q;++i) {
		scanf("%d",&op);
		if(op==1) {
			scanf("%d",&l);
			c=getchar();while(c<'a'||c>'z') c=getchar();
			if(s[l]==c) continue ;
			modify(-1,l,s[l]-'a');
			s[l]=c;
			modify(1,l,s[l]-'a');
		}
		else {
			ans=0;
			scanf("%d%d",&l,&r);
			for(int j=0;j<26;++j) if(query(r,j)-query(l-1,j)>0) ans++;
			printf("%d\n",ans);
		}
	}
}