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;
	char s[500010];
	for(ll i=0;i<n;i++){
		char now;
		cin>>now;
		s[i]=now;
	}
	set<ll>ch[26];
	for(ll i=0;i<n;i++){
		char now=s[i];
		ch[now-'a'].insert(i);
	}
	ll q;cin>>q;
	for(ll i=0;i<q;i++){
		ll f;
		cin>>f;
		if(f==2){
			ll cnt=0;
			ll l,r;
			cin>>l>>r;
			l--;
			for(ll j=0;j<26;j++){
				auto itr1=ch[j].lower_bound(l);
				auto itr2=ch[j].lower_bound(r);
				if(itr1!=itr2){
					cnt++;
				}
			}
			if(l==r){
				cnt++;
			}
			cout << cnt<<endl;
		}
		if(f==1){
			ll now;char t;
			cin>>now;
			now--;cin>>t;
			ch[s[now]-'a'].erase(now);
			s[now]=t;
			ch[t-'a'].insert(now);
		}
	}
	// your code goes here
	return 0;
}