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>
#define rep(i,n) for(int i = 0; i < n; i++)
#define pb push_back
using namespace std;
typedef long long ll;

int main(){
  int n;
  cin>>n;
  string s;
  cin>>s;
  int q;
  cin>>q;

  vector<set<int>> d(26,set<int>({1000000}));
  rep(i,n) d[s[i]-'a'].insert(i);

  rep(num,q){
    int type;
    cin>>type;
    if(type==1){
      int i;
      char c;
      cin>>i>>c;
      i--;
      d[s[i]-'a'].erase(i);
      d[c-'a'].insert(i);
      s[i]=c;
    }else{
      int l,r;
      cin>>l>>r;
      l--;
      int res=0;
      rep(i,26){
        int x,y;
        if(d[i].empty()) continue;
        x=*d[i].lower_bound(l);
        y=*d[i].lower_bound(r);
        if(y>x) res++;
      }
      cout<<res<<endl;
    }
  }
}
