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,i,j,t,l,r,cnt;
  string s;
  char c;
  cin>>n>>s>>q;
  vector<set<int>> a(26);
  for(i=0;i<n;i++) a.at(s.at(i)-'a').insert(i+1);
  for(j=1;j<=q;j++) {
    cin>>t;
    if(t==1) {
      cin>>i>>c;
      if (s.at(i-1)!=c) {
        a.at(s.at(i-1)-'a').erase(i);
        a.at(c-'a').insert(i);
        s.at(i-1)=c;
      }
    }
    else if(t==2) {
      cin>>l>>r;
      cnt=0;
      for(i=0;i<26;i++){
        auto m=a.at(i).lower_bound(l);
        if(m!=a.at(i).end() && *m<=r) cnt++;
      }
      cout<<cnt<<endl;
    }
  }
}