#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)
using ll=long long;
const ll MOD=1000000007;


int main() {
    int N,Q;
    string S;
    cin>>N>>S>>Q;
    vector<set<int>> atoz(26);
    rep(i,N){
        atoz.at(S.at(i)-'a').insert(i);
    }
    
    rep(i,Q){
        int a;
        cin>>a;
        if(a==1){
            int b;
            char c;
            cin>>b>>c;--b;
            atoz.at(S[b]-'a').erase(b);
            S[b]=c;
            atoz.at(c-'a').insert(b);
        }
        if(a==2){
            int b,c;
            cin>>b>>c;--b;
            int count=0;
            rep(j,26){
                auto p=atoz[j].lower_bound(b);
                if(p!=atoz[j].end() && *p<c){count++;}
            }
            cout<<count<<endl;
        }
    }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)
using ll=long long;
const ll MOD=1000000007;


int main() {
    int N,Q;
    string S;
    cin>>N>>S>>Q;
    vector<set<int>> atoz(26);
    rep(i,N){
        atoz.at(S.at(i)-'a').insert(i);
    }
    
    rep(i,Q){
        int a;
        cin>>a;
        if(a==1){
            int b;
            char c;
            cin>>b>>c;--b;
            atoz.at(S[b]-'a').erase(b);
            S[b]=c;
            atoz.at(c-'a').insert(b);
        }
        if(a==2){
            int b,c;
            cin>>b>>c;--b;
            int count=0;
            rep(j,26){
                auto p=atoz[j].lower_bound(b);
                if(p!=atoz[j].end() && *p<c){count++;}
            }
            cout<<count<<endl;
        }
    }
}

```