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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
  int n; string s; int q; cin >> n >> s >> q;
  vector<set<int>> S(26);
  rep(i, n) S[s[i] - 'a'].insert(i);
  rep(i, q) {
    int a; cin >> a;
    if (a == 1) {
      int b; char c; cin >> b >> c; b--;
      S[s[b] - 'a'].erase(b);
      s[b] = c;
      S[s[b] - 'a'].insert(b);
    }
    else if (a == 2) {
      int l, r; cin >> l >> r; l--;
      int cnt = 0;
      rep(j, 26) {
        auto it = S[j].lower_bound(l);
        if (it != S[j].end() && *it < r) cnt++;
      }
      cout << cnt << endl;
    }
  }
  return 0;
}