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

int main(){
  int N,Q;
  string S;
  cin >> N >> S >> Q;
  vector<set<int>> A(26);
  rep(i,N){
    A.at(S.at(i)-'a').insert(i+1);
  }
  rep(i,Q){
    int t;
    cin >> t;
    if(t==1){
      ll j;
      char c;
      cin >> j >> c;
      A.at(S.at(j-1)-'a').erase(j);
      S.at(j-1)=c;
      A.at(c-'a').insert(j);
    }else{
      int l,r,ans=0;
      cin >> l >> r;
      rep(j,26){
        auto it=A.at(j).lower_bound(l);
        if(A.at(j).end()!=it&&*it<=r) ans++;
      }
      cout << ans << endl;
    }
  }    
}

### Prompt

Create a solution in CPP for the following problem:

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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define all(v) v.begin(), v.end()
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
const ll MOD =1000000007;

int main(){
  int N,Q;
  string S;
  cin >> N >> S >> Q;
  vector<set<int>> A(26);
  rep(i,N){
    A.at(S.at(i)-'a').insert(i+1);
  }
  rep(i,Q){
    int t;
    cin >> t;
    if(t==1){
      ll j;
      char c;
      cin >> j >> c;
      A.at(S.at(j-1)-'a').erase(j);
      S.at(j-1)=c;
      A.at(c-'a').insert(j);
    }else{
      int l,r,ans=0;
      cin >> l >> r;
      rep(j,26){
        auto it=A.at(j).lower_bound(l);
        if(A.at(j).end()!=it&&*it<=r) ans++;
      }
      cout << ans << endl;
    }
  }    
}

```