// Om Namah Shivaya
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)
template<typename T>
void amin(T &a, T b) {
a = min(a, b);
}
template<typename T>
void amax(T &a, T b) {
a = max(a, b);
}
#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif
/*
ig i read the edi for this problem a long time ago (dont remember the approach in the edi tho)
greedy strategy
=> in each op, try to reduce the number as much as possible
because the #of ops is a montonic sequence
if we need k ops to make n1 = 0, then we would need <= k ops to make n2 = 0, n2 < n1
because we can apply the same k ops we applied to n1 to n2 and n2 would also become 0
i remember seeing this idea somewhere => maybe in the edi of this problem/some other problem
*/
const int MOD = 1e9 + 7;
const int N = 1e7 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
vector<int> spf(N);
void precalc() {
rep(i, N) spf[i] = i;
for (int i = 2; i * i < N; ++i) {
if (spf[i] != i) conts;
for (int j = i * i; j < N; j += i) {
amin(spf[j], i);
}
}
}
void solve(int test_case)
{
precalc();
int n, q; cin >> n >> q;
vector<int> a(n);
rep(i, n) cin >> a[i];
vector<bool> there(N);
rep(i, n) there[a[i]] = 1;
vector<int> cnt(N);
cnt[0] = n;
vector<ll> dp(N, inf1);
dp[0] = 0;
int ptr = 0;
rep1(i, N - 1) {
int x = i;
int prevp = -1;
while (x > 1) {
int p = spf[x];
if (there[p] and p != prevp) {
cnt[i - p]--;
cnt[i]++;
}
x /= p;
prevp = p;
}
while (ptr < N - 1 and !cnt[ptr]) {
ptr++;
}
dp[i] = dp[ptr] + 1;
}
while (q--) {
int m; cin >> m;
if (dp[m] < inf1) cout << dp[m] << endl;
else cout << "oo" << endl;
}
}
int main()
{
fastio;
int t = 1;
// cin >> t;
rep1(i, t) {
solve(i);
}
return 0;
}
