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<iostream>
#include<string>
#include<vector>
#include<set>

using namespace std;

int main(){
    int siz;
    scanf(" %d",&siz);
	string re;
	cin >> re;
	vector<set<int>> let(26);
	for(int i = 0; i < int(re.size());++i)
		let[re[i]-'a'].insert(i);

	int q;
	scanf(" %d",&q);
	while(q--){
		int t;
		scanf(" %d",&t);
		if(t == 1){
			int pl;
			scanf(" %d",&pl);
			--pl;
			char add;
			scanf(" %c",&add);
			let[re[pl] - 'a'].erase(pl);
			re[pl] = add;
			let[re[pl]-'a'].insert(pl);
		}
		else{
			int ans = 0;
			int l,r;
			scanf(" %d %d",&l,&r);
			--l,--r;
			for(int i = 0; i < 26;++i){
				auto it = let[i].lower_bound(l);
				if(it != let[i].end() && *it <= r) ++ans;
			}
			printf("%d\n",ans);
		}
	}
	return 0;
}
