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 N 500005
int n,m;char a[N];
struct BIT
{
	int v[N];
	void upd(int x,int _v) {for(;x<=n;x+=x&-x) v[x]+=_v;}
	int qSm(int x) {int res=0;for(;x;x-=x&-x) res+=v[x];return res;} 
}BT[26];
int main()
{
	scanf("%d %s",&n,a+1);
	for(int i=1;i<=n;++i) BT[a[i]-'a'].upd(i,1);scanf("%d",&m);
	for(int i=1,op,l,r,x,res;i<=m;++i)
	{
		char c[2];scanf("%d",&op);
		if(op==1) scanf("%d %s",&x,c),BT[a[x]-'a'].upd(x,-1),BT[c[0]-'a'].upd(x,1),a[x]=c[0];
		if(op==2)
		{
			res=0;scanf("%d %d",&l,&r);
			for(int j=0;j<26;++j) if(BT[j].qSm(r)>BT[j].qSm(l-1)) ++res;printf("%d\n",res);
		}
	}return 0;
}