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>
#include <string>
#include <vector>
#include <algorithm>
#define rep(i,w) for (int i = 0;i < (w); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;
int INF = 1e9;

int main(){
  int n,q;
  string s;
  cin >> n >> s >> q;
  vector<set<int>> table(26);
  rep(i,s.size()) table[s[i]-'a'].insert(i);
  rep(query,q){
    int com;cin >> com;
    if(com == 1){
      int i;
      char ch;
      cin >> i >> ch;
      --i;
      table[s[i]-'a'].erase(i);
      s[i] = ch;
      table[ch-'a'].insert(i);
    }else{
      int ans = 0;
      int l,r;cin >> l >> r;
      --r;--l;
      rep(j,26){
        set<int>::iterator it = table[j].lower_bound(l);
        if(it != table[j].end() && *it <= r) ans++;
      }
      printf("%d\n",ans);
    }
  }
}