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;
typedef double dd;
const int N = 1e5+5;
const dd eps = 1e-8;

vector<set<int>> st(27);
string now,c;
int qu[500005];
int m,n,q,l,r;
int main(){
  ios::sync_with_stdio(false);
  // freopen("input.txt","r",stdin);
  cin>>n>>now;
  for(int i = 1;i <= n;i++){    
    qu[i] = now[i-1] - 'a';
    st[qu[i]].insert(i);
  }
  cin>>m;
  while(m--){
    cin>>q;
    if(q == 1){
      cin>>l>>c;
      st[qu[l]].erase(l);
      qu[l] = c[0] - 'a';
      st[qu[l]].insert(l);
    }
    else{
      cin>>l>>r;
      int cnt = 0;
      for(int i = 0;i < 26;i++){
	if(st[i].lower_bound(l) != st[i].lower_bound(r+1)) cnt++;
      }
      cout<<cnt<<endl;
    }
  }
  return 0;
}
