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 <iostream>
#include <algorithm>
#include <vector>
#include <set>

using namespace std;

const int INF = 1000000009;

int main() {
  int n; cin >> n;
  string s; cin >> s;

  vector<set<int>> a(26, set<int>({INF})); // INF は番兵のようなもの
  for (int i = 0; i < n; i++) { a[s[i]-'a'].insert(i); }

  int q; cin >> q;
  while (q--) {
    int type; cin >> type;
    if (type == 1) {
      int idx; char c; cin >> idx >> c; idx--;
      a[s[idx]-'a'].erase(idx);
      a[c-'a'].insert(idx);
      s[idx] = c;
    }
    if (type == 2) {
      int l, r; cin >> l >> r; l--;
      int ret = 0;
      for (int i = 0; i < 26; i++) {
        int lid = *a[i].lower_bound(l);
        if (lid < r) ret++;
      }
      cout << ret << endl;
    }
  }
  return 0;
}