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+20,C=26;
char a[N];
int n,q,type,pos,l,r;
char c;
set <int> s[C];
int main()
{
	scanf("%d %s%d",&n,a+1,&q);
	for(int i=1;i<=n;i++) s[a[i]-'a'].insert(i);

	while(q--)
	{
		scanf("%d",&type);
		if(type==1)
		{
			scanf("%d %c",&pos,&c);
			s[a[pos]-'a'].erase(pos);
			s[c-'a'].insert(pos);
			a[pos]=c;
		}
		else
		{
			scanf("%d%d",&l,&r);

			int ans=0;
			for(int i=0;i<C;i++)
			{
				auto it=s[i].lower_bound(l);
				if(it==s[i].end() or *it>r) continue;
				ans++;
			}
			printf("%d\n",ans);
		}
	}
}