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;
#define lowbit(x) x&-x
	int n,m;
struct fenwick
{
	int c[1000001];
	void add(int x,int ind)
	{
		while(x<=n)
			c[x]+=ind,x+=lowbit(x);
		return;
	}
	int ask(int x)
	{
		int ans=0;
		while(x)
			ans+=c[x],x-=lowbit(x);
		return ans;
	}
}f[27];
char s[1000001];
signed main() 
{
	scanf("%d",&n);
	cin>>s+1;
	cin>>m;
	for(int i=1;i<=n;i++)
		f[s[i]-'a'].add(i,1);
	for(int i=1;i<=m;i++)
	{
		int op,l,r;
		scanf("%d",&op);
		if(op==1)
		{
			char c;
			cin>>l>>c;
			f[s[l]-'a'].add(l,-1);
			s[l]=c;
			f[s[l]-'a'].add(l,1);
		}
		else
		{
			int ans=0;
			scanf("%d%d",&l,&r);
			for(int j=0;j<26;j++)
				if(f[j].ask(r)-f[j].ask(l-1))
					ans++;
			printf("%d\n",ans);
		}
	}
	return 0;
}
/*
5
1 2 4 9 10
4
1 2
1 3
1 4
1 5
*/