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() {
  array<set<int>, 26> letters;
  int n, q, t, x, y;
  cin >> n;
  char c, s[n];
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
    letters[s[i] - 'a'].insert(i);
  }
  cin >> q;
  while (q--) {
    cin >> t >> x;
    if (t == 1) {
      // x index, c char
      cin >> c;
      if (s[x-1] == c) continue;
      letters[s[x-1] - 'a'].erase(x-1);
      letters[c - 'a'].insert(x-1);
      s[x-1] = c;
    } else {
      cin >> y; // x:left, y: right
      int ans(0);
      for (int i = 0; i < 26; ++i) {
        auto it = letters[i].lower_bound(x-1);
        if (it != letters[i].end() && *it < y) ans++;
      }
      cout << ans << '\n';
    }
  } 
  
  return 0;
}