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 LL;
const int N = 5e5 + 10;
#define fi first
#define se second
#define pb push_back
int t[26][N];
int n,q;
char a[N];
void add(int x,int p,int d){
  for(;p<N;p+=p&-p)t[x][p]+=d;
}
int get(int x,int p){
  int d=0;
  for(;p;p-=p&-p)d+=t[x][p];
  return d;
}
int main() {
  ios::sync_with_stdio(false);
  cin>>n>>a+1>>q;
  for(int i=1;i<=n;i++){
    add(a[i]-'a',i,1);
  }
  for(int i=1;i<=q;i++){
    int o;
    cin>>o;
    if(o==1){
      int p;char d;
      cin>>p>>d;
      add(a[p]-'a',p,-1);
      a[p]=d;
      add(a[p]-'a',p,1);
    }else{
      int l,r,ans=0;
      cin>>l>>r;
      for(int j=0;j<26;j++){
        if(get(j,r)-get(j,l-1))ans++;
      }
      cout<<ans<<'\n';
    }
  }
  return 0;
}