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;
  cin>>n;
  string t;
  cin>>t;
  char s[n];
  for (int i=0; i<n; i++) s[i] = t[i];
  int q;
  cin>>q;
  set<int> m[26];
  for (int i=0; i<n; i++) {
    m[s[i]-'a'].insert(i);
  }
  for (int _=0; _<q; _++) {
    int type;
    cin>>type;
    if (type == 1) {
      int i; char c;
      cin>>i>>c;
      --i;
      char prev = s[i];
      m[prev-'a'].erase(i);
      m[c-'a'].insert(i);
      s[i] = c;
      continue;
    }
    int l, r;
    cin>>l>>r;
    --l; --r;
    int cnt = 0;
    for (int i=0; i<26; i++) {
      char c = 'a' + i;
      auto it = m[c-'a'].lower_bound(l);
      if (it != m[c-'a'].end()) {
        if (*it <= r) {
          ++cnt;
        }
      }
    }
    cout<<cnt<<endl;
  }
}