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<bits/stdc++.h>
using namespace std;
set<int> se[26];
int N,n,m,a,b,d,l,r;
char c;
string s;
int main(void){
    cin>>N>>s>>n;
    for(int i=0;i<s.size();i++){
        se[s[i]-'a'].insert(i);
    }
    for(int i=0;i<n;i++){
        int ans=0;
        cin>>a>>b;
        b--;
        if(a==1)cin>>c;
        else {cin>>d;d--;}
        if(a==1){
            se[s[b]-'a'].erase(b);
            se[c-'a'].insert(b);
            s[b]=c;
        }
        else{
            for(int i=0;i<26;i++){
                auto itr1=se[i].upper_bound(d);
                auto itr2=se[i].lower_bound(b);
                if(itr1!=itr2)ans++;
            }
            cout<<ans<<endl;
        }
        //cout<<s<<endl;
    }
}
