#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <sstream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <numeric>
#include <cctype>
#include <tuple>
#include <climits>
#include <bitset>
#include <cassert>
#include <random>
#include <complex>

#ifdef _MSC_VER
#include <agents.h>
#endif

#define FOR(i, a, b) for(int i = (a); i < (int)(b); ++i)
#define rep(i, n) FOR(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define REV(v) v.rbegin(), v.rend()
#define MEMSET(v, s) memset(v, s, sizeof(v))
#define UNIQUE(v) (v).erase(unique(ALL(v)), (v).end())
#define MP make_pair
#define MT make_tuple

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

unsigned xor128_x = 123456789, xor128_y = 362436069, xor128_z = 521288629, xor128_w = 88675123;
unsigned xor128() {
	unsigned t = xor128_x ^ (xor128_x << 11);
	xor128_x = xor128_y; xor128_y = xor128_z; xor128_z = xor128_w;
	return xor128_w = xor128_w ^ (xor128_w >> 19) ^ (t ^ (t >> 8));
}

void generateA(int N, vector<int> &A) {
	for (int i = 0; i < N; ++i)
		A[i] = xor128() % 100003;
}

const int mod = 100003;

ll mod_pow(ll x, ll n){
	ll res = 1;
	while (n){
		if (n & 1) res = res*x%mod;
		x = x*x%mod;
		n /= 2;
	}
	return res;
}

ll calcrev(ll x){
	return mod_pow(x, mod - 2);
}

int rev[mod];

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n, q;
	cin >> n >> q;


	if (n <= 1000){
		vector<int> a(n);
		generateA(n, a);
		while (q--){
			ll x;
			cin >> x;
			ll ans = 0;
			rep(i, n) ans = max(ans, a[i] * x%mod);
			cout << ans << '\n';
		}
	}
	else{
		vector<int> b(n);
		generateA(n, b);

		vector<int> a(mod);
		for (auto e : b) a[e] = 1;

		for (int i = 1; i < mod; ++i) rev[i] = calcrev(i);
		int cnt = 0;
		vector<int> ans(mod);
		rep(i, mod){
			for (ll j = mod - 1; j >= 0; --j){
				++cnt;
				if (a[j*rev[i] % mod]){
					ans[i] = j;
					break;
				}
			}
		}
		ans[0] = 0;
		while (q--){
			int x;
			cin >> x;
			cout << ans[x] << '\n';
		}
	}



	return 0;
}