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>
#define rep(i,n) for(int i=0; i<(int)(n); i++)

using namespace std;
typedef long long LL;

int main(){
	int N;
	cin >> N;
	string S;
	cin >> S;
	int Q;
	cin >> Q;
	set<int> s[26];
	char base='a';
	rep(i,N){
		int num=S[i]-base;
		s[num].insert(i);
	}
	while(Q--){
		int t;
		cin >> t;
		if(t==1){
			int i;
			char c;
			cin >> i >> c;
			i--;
			int del, add=c-base;
			rep(j,26){
				auto itr=s[j].find(i);
				if(itr==s[j].end()) continue;
				del=j;
				break;
			}
			s[del].erase(i);
			s[add].insert(i);
		}
		if(t==2){
			int l, r;
			cin >> l >> r;
			l--; r--;
			int ans=0;
			rep(i,26){
				auto itr=s[i].lower_bound(l);
				if(itr!=s[i].end() && *itr<=r) ans++;
			}
			cout << ans << endl;
		}
	}

	return 0;
}