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;
#define INF 1LL<<62
#define inf 1000000007

int main() {
	ll n;
	cin>>n;
	string s;
	cin>>s;
	ll q;
	cin>>q;
	set<ll>ch[26];
	for(ll i=0;i<n;i++){
		ll now=s[i]-'a';
		//cout <<now;
		ch[now].insert(i);
	}
	//cout <<endl;
	for(ll i=0;i<q;i++){
		ll y;
		cin>>y;
		if(y==1){
			ll x;char t;
			cin>>x>>t;
			x--;
			ll now=s[x]-'a';
			ch[now].erase(x);
			ll nex=t-'a';
			ch[nex].insert(x);
			s[x]=t;
			//cout <<s<<endl;
			//cout <<nex<<endl;
		}
		else{
			ll l,r;
			cin>>l>>r;
			l--;r--;
			ll cnt=0;
			for(ll j=0;j<26;j++){
				auto itr1=ch[j].lower_bound(l);
				ll la=*itr1;
				if(itr1==ch[j].end()){
					la=inf;
				}
				if(la<=r){
					cnt++;
				}
			}
			cout <<cnt<<endl;
		}
	}
	// your code goes here
	return 0;
}