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, t, l, r, a;
  string S;
  set<int> s[26];
  char c;
  cin >> N;
  cin >> S;
  for (int i=0; i<N; i++)
    s[S[i] - 'a'].insert(i);
  cin >> Q;
  for (int i=0; i<Q; i++) {
    cin >> t;
    if (t == 1) {
      cin >> a >> c;
      a--;
      s[S[a]-'a'].erase(a);
      s[c-'a'].insert(a);
      S[a] = c;
    } else {
      cin >> l >> r;
      l--;
      r--;
      int cnt = 0;
      for (int j=0; j<26; j++) {
        auto it = s[j].lower_bound(l);
        if (it == s[j].end() || *it > r)
          continue;
        cnt++;
      }
      cout << cnt << endl;
    }
  }
  return 0;
}