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;

int main(){
  int n, q;
  string s;
  cin >> n >> s >> q;
  vector<set<int>> arf(26);
  for(int i=0; i<n; ++i){
    arf[s[i]-'a'].insert(i);
  }
  for(int i=0; i<q; ++i){
    int q1;
    cin >> q1;
    if(q1 == 1){
      int q2;
      char q3;
      cin >> q2 >> q3;
      --q2;
      arf[s[q2]-'a'].erase(q2);
      s[q2] = q3;
      arf[q3-'a'].insert(q2);
    }else{
      int q2, q3;
      cin >> q2 >> q3;
      --q2;
      int ans = 0;
      for(int j=0; j<26; ++j){
        auto it = arf[j].lower_bound(q2);
        if(it != arf[j].end() && *it < q3) ++ans;
      }
      printf("%d\n",ans);
    }
  }
}
