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;

typedef long long int ll;

const ll mod = 1e9 + 7;

int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(0);cout.tie(0);

  ll n;
  cin >> n;
  string s;
  cin >> s;
  vector < set < ll > > v(26);
  for(int i = 0; i < n; i++){
    v[s[i] - 'a'].insert(i + 1);
  }

  ll q;
  cin >> q;
  while(q--){
    int a;
    cin >> a;
    if(a == 1){
      ll pos;
      cin >> pos;
      char ch;
      cin >> ch;
      v[s[pos - 1] - 'a'].erase(pos);
      v[ch - 'a'].insert(pos);
      s[pos - 1] = ch;
    }else{
      ll l, r;
      cin >> l >> r;
      ll cnt = 0;
      for(int i = 0; i < 26; i++){
        auto it = v[i].lower_bound(l);
        if(it != v[i].end() && *it <= r) cnt++;
      }
      cout << cnt << '\n';
    }
  }
}
