/*
When doing hashing:
- pick a key big enough to make sure you won't have too many collisions
    - you can pick your key about 6 times bigger than the number of elements from the hash
- pick a prime number key
- pick a key small enough to make sure you won't waste too much memory on the list
*/

/*
You are given an array of N elements.
You are also given M querys - each query has a value X
For each X output "True" if X is in the initial array, or "False" if it isn't
*/

/*
    FIRST SOLUTION:
    binary search
    - we should sort the N elements array (merge sort)
    - then binary search for each of the M elements in the original array.
    - Complexity: O(NlogN + MlogN) = O(logN * (N + M))
*/

/*
n = 5, m = 7
v[] = 1 2 3 4 5
3 - True
4 - True
5 - True
2 - True
9 - False
6 - False
4 - True
*/

/*
    SECOND SOLUTION:
    with hashing
    - read the N elements and insert all of them in your hash
    - read the M queryes and look up for each of them in your hash
    - Complexity: O(N + M)
*/
#include <iostream>
#include <vector>
using namespace std;

// total memory - KEY + N
const int KEY = 666013;

vector<int> H[KEY];
void insert(int x) { // insert in the hash - O(1)
    int line = x % KEY;
    H[line].push_back(x);
}
bool query(int x) {
    int line = x % KEY;
    for (size_t i = 0; i < H[line].size(); ++i) {
        if (H[line][i] == x) {
            return true;
        }
    }
    return false;
}

int n, x;
int m;

int main() {
    cin >> n >> m; // read the number of elements in your array and the number of querys
    for (int i = 0; i < n; ++i) {
        cin >> x; // read your x
        insert(x); // insert your number x in the hash
    }
    for (int i = 0; i < m; ++i) { // for each of the m querys
        cin >> x;
        if (query(x) == true) {
            cout << "Number " << x << " is in the hash\n";
        } else {
            cout << "Number " << x << " is not in the hash\n";
        }
    }
}