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;

using ll = long long;
const int N = 5e5+10;

int n, q;
char s[N];
set<int> idx[26];

int main () {
  scanf("%d %s %d", &n, s + 1, &q);
  for(int i = 1; i <= n; i++)
    idx[s[i] - 'a'].insert(i);
  while(q--) {
    int ty; scanf("%d", &ty);
    if(ty == 1) {
      int i; char x;
      scanf(" %d %c", &i, &x);

      idx[s[i] - 'a'].erase(i);
      s[i] = x;
      idx[s[i] - 'a'].insert(i);
    } else {
      int l, r; scanf(" %d %d", &l, &r);
      int ans = 0;
      for(int i = 0; i < 26; i++) {
        auto it = idx[i].lower_bound(l);
        if(it != idx[i].end() && *it <= r) ans++;
      }
      printf("%d\n", ans);
    }
  } return 0;
}