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

int main(){
  int n;
  cin >> n;
  string s;
  cin >> s;
  int q;
  cin >> q;
  vector<set<int>> count(26);
  rep(i, n) {
    count[s[i] - 'a'].insert(i);
  }
  rep(qi, q){
    int type;
    cin >> type;
    if(type == 1){
      int i; char c;
      cin >> i >> c;
      i--;
      count[s[i] - 'a'].erase(i);
      s[i] = c;
      count[c - 'a'].insert(i);
    } else {
      int l, r;
      cin >> l >> r;
      l--; r--;
      int ans = 0;
      rep(i, 26){
        auto it = count[i].lower_bound(l);
        if(it != count[i].end() && *it <= r) ans++;
      }
      cout << ans << endl;
    }
  }
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;
typedef long long ll;
typedef pair<int,int> P;
#define rep(i,N) for(ll (i)=0;(i)<(N);(i)++)
const int mod = 1000000007;

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