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 <vector>
using namespace std;
// https://atcoder.jp/contests/abc157/submissions/10486180

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N,Q;
  string S;
  cin >> N >> S >> Q;
  vector<int> A(N);
  for(int i=0;i<N;++i){
    A[i] = 1 << (S[i]-'a');
  }

  int x,l,r,i;
  char c;
  while(Q--){
    cin >> x;
    if(x==1){
      cin >> i >> c;
      i--;
      A[i] = 1<<(c-'a');
    }else{
      cin >> l >> r;
      int ans = 0;
      for(int i=l-1;i<r;i++)ans|=A[i];
      cout << __builtin_popcount(ans) << '\n';
    }
  }
}