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;
#define rep(i, a) for(int i=0; i<(a); i++)
typedef long long ll;

int main()
{
  int n; cin>>n;
  vector<set<int>> vst(26);
  string s; cin>>s;
  for (int i=0; i<(int)(s.size()); i++) {
    vst[s[i]-'a'].insert(i);
  }
  int q; cin>>q;
  vector<int> ans;
  rep(i, q) {
    int a; cin >> a;
    if (a==1) {
      int b; cin >> b; b--;
      char c; cin >> c;
      rep(j, 26) vst[j].erase(b);
      vst[c-'a'].insert(b);
    } else {
      int b, c; cin>>b>>c; b--; c--;
      int sum = 0;
      rep(i, 26) {
        auto itr = vst[i].lower_bound(b);
        if (itr == vst[i].end()) continue;
        if (*itr <= c) sum++;
      }
      ans.push_back(sum);
    }
  }
  for (auto v:ans) cout << v << endl;

  return 0;
}
