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>> t(26);
  for (int i = 0; i < n; i++) {
    t.at(s.at(i) - 'a').insert(i);
  }
  for (int i = 0; i < q; i++) {
    int type;
    cin >> type;
    if (type == 1) {
      int iq;
      char cq;
      cin >> iq >> cq;
      iq--;
      t.at(s.at(iq) - 'a').erase(iq);
      s.at(iq) = cq;
      t.at(cq - 'a').insert(iq);
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      int ans = 0;
      for (char c = 'a'; c <= 'z'; c++) {
        if (t.at(c - 'a').lower_bound(l)
           != t.at(c - 'a').lower_bound(r)) ans++;
      }
      cout << ans << endl;
    }
  }
}
