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 <iostream>
#include <string>
#include <set>
#include <algorithm>
using namespace std;
const int INF = 1e8;
int N, Q; string S; set<int> s[26];
int main() {
  cin >> N >> S >> Q;
  for(int i = 0; i < 26; ++i) s[i].insert(INF);
  for(int i = 0; i < N; ++i) s[S[i]-'a'].insert(i+1);
  for(int i = 0; i < Q; ++i) {
    int type, iq, lq, rq; char cq;
    cin >> type;
    if(type == 1) {
      cin >> iq >> cq;
      s[S[iq-1]-'a'].erase(iq);
      s[cq-'a'].insert(iq);
      S[iq-1] = cq;
    } else {
      cin >> lq >> rq;
      int cnt = 0;
      for(int j = 0; j < 26; ++j) {
        auto itrL = s[j].lower_bound(lq);
        auto itrR = s[j].upper_bound(rq);       
        if(*itrR > *itrL) ++cnt;
      }
      cout << cnt << endl;
    }
  }
}