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;
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
typedef long long ll;

int main(){
	int n,q;
	string s;
	cin >> n >> s >> q;
	vector<set<int>> S(26);
	rep(i,n){
		S[s[i]-'a'].insert(i);
	}
	rep(iq,q){
		int q; cin >> q;
		if(q == 1){
			int i;
			char c;
			cin >> i >> c;
			i--;
			S[s[i] - 'a'].erase(i);
			s[i] = c;
			S[s[i] - 'a'].insert(i);
		} else if (q == 2){
			int l,r;
			cin >> l >> r;
			l--;r--;
			int ans = 0;
			rep(j,26){
				auto itr = S[j].lower_bound(l);
				if(itr != S[j].end() && *itr <= r) ans++;
			}
			cout << ans << endl;
		}
	}
    return 0;
}

