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 ll long long
using namespace std;

int main() {
  int n; cin>>n;
  string S; cin>>S;
  vector<set<int>> B(26);
  rep(i, n) B[S[i]-'a'].insert(i);
  int Q; cin>>Q;
  vector<int> ans;
  rep(query, Q) {
    int type; cin>>type;
    if (type==1) {
      int i; cin>>i;
      char c; cin>>c;
      i--;
      B[S[i]-'a'].erase(i);
      S[i]=c;
      B[S[i]-'a'].insert(i);
    }
    else {
      int l,r; cin>>l>>r;
      l--; r--;
      int res=0;
      rep(i, 26) {
        auto itr=B[i].lower_bound(l);
        if (itr!=B[i].end()&&*itr<=r) res++;
      }
      ans.push_back(res);
    }
  }
  rep(i, (int)(ans.size())) cout<<ans[i]<<endl;
}
