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;
using u32 = uint32_t;

int main() {
	u32 N;
	cin >> N;
	
	string S;
	cin >> S;
	
	u32 Q;
	cin >> Q;
	
	map<char, set<u32>> cm;
	for(u32 i=0; i<N; i++) cm[S[i]].insert(i);

	for(u32 q=0; q<Q; q++) {
		u32 t;
		cin >> t;
		
		if(t & 1) {
			u32 i;
			char c;
			cin >> i >> c;
	
			if(S[i-1] == c) continue;
			
			cm[c].insert(i-1);
			cm[S[i-1]].erase(i-1);
			
			if(cm[S[i-1]].empty()) cm.erase(S[i-1]);
			
			S[i-1] = c;
			
		} else {
			u32 l, r;
			cin >> l >> r;
			
			u32 ans = 0;
			
			for(auto ci=begin(cm); ci!=end(cm); ci++) {
				auto pi = (*ci).second.lower_bound(l - 1);
				if(pi != (*ci).second.end() && *pi <= r - 1) ans++;
			}
			
			cout << ans << endl;
		}
	}
	
	return 0;
}