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;
typedef long long ll;
const int maxn=5e5+100;
char ch[maxn];
int n,arr[27][maxn];
int lowbit(int n)
{
	return n&-n;
}
void update(int p,int t,int a[])
{
	for(int i=p;i<=n;i+=lowbit(i))
		a[i]+=t;
}
ll getsum(int p,int a[])
{
	ll ans=0;
	for(int i=p;i>0;i-=lowbit(i))
		ans+=a[i];
	return ans;
}
int main()
{
	int m;
	cin>>n;
	scanf("%s",ch+1);
	for(int i=1;i<=n;i++)
		update(i,1,arr[ch[i]-'a']);
	cin>>m;
	for(int i=1;i<=m;i++)
	{
		int t;
		cin>>t;
		if(t==1)
		{
			char c;
			int p;
			cin>>p>>c;
			update(p,-1,arr[ch[p]-'a']);
			ch[p]=c;
			update(p,1,arr[c-'a']);
		}
		if(t==2)
		{
			ll sum=0,x,y;
			cin>>x>>y;
			for(int i=0;i<26;i++)
				sum+=getsum(y,arr[i])-getsum(x-1,arr[i])>0;
			cout<<sum<<endl;
		}
	}
    return 0;
} 
