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;
typedef long long ll;

int main()
{
  int n, q;
  string s;
  cin >> n >> s >> q;
  vector<set<int>> is(26);
  rep(i, n)
    is[s[i] - 'a'].insert(i);
  rep(qi, q)
  {
    int ty;
    cin >> ty;
    if (ty == 1)
    {
      int i; char c;
      cin >> i >> c;
      i--;
      is[s[i] - 'a'].erase(i);
      s[i] = c;
      is[s[i] - 'a'].insert(i);
    }
    else
    {
      int l, r;
      cin >> l >> r;
      l--;
      int count = 0;
      rep(i, 26)
      {
        auto it = is[i].lower_bound(l);
        if (it != is[i].end() && *it < r)
          count++;
      }
      printf("%d\n", count);
    }
  }
}