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() {
  ios::sync_with_stdio(false);
  int N, Q;
  string S;
  cin >> N >> S >> Q;
  vector<set<int>> V(26);
  for (int i = 0; i < N; i++) V.at(S.at(i) - 'a').insert(i + 1);
  while (Q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int i;
      char c;
      cin >> i >> c;
      V.at(S.at(i - 1) - 'a').erase(i);
      S.at(i - 1) = c;
      V.at(S.at(i - 1) - 'a').insert(i);
    } else {
      int l, r;
      cin >> l >> r;
      int ans = 0;
      for (int i = 0; i < 26; i++) {
        auto iter = V.at(i).lower_bound(l);
        if (iter != V.at(i).end() && *iter <= r) ans++;
      }
      cout << ans << "\n";
    }
  }
}