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;
int main() {
  int n, q;
  string s;
  cin >> n >> s >> q;
  vector<set<int>> se(26);
  for (int i = 0; i < s.size(); i++) {
    se[s[i] - 'a'].insert(i);
  }
  for (int i = 0; i < q; i++) {
    int type;
    cin >> type;
    if (type == 1) { //puts("hoge");
      int a; char b;
      cin >> a >> b; --a;
      auto lb = se[s[a] - 'a'].find(a) ;
      if (lb != se[s[a] - 'a'].end()) se[s[a] - 'a'].erase(lb);
      se[b - 'a'].insert(a);
      s[a] = b;
    }
    else { //puts("fuga");
      int a, b;
      cin >> a >> b; --a, --b;
      int cnt = 0;
      for(int c = 0; c < 26; c++) {
        auto it = se[c].lower_bound(a);
        cnt += (it != se[c].end() && *it <= b);
      }
      cout << cnt << "\n";
    }
  }
}