/*
There is a simple solution with upper bound on the time equal to O(MAX * log(log(MAX)) * log(MAX)) with DP and maintaining a heap. Probably the actual complexity is smaller.
*/
#include <bits/stdc++.h>
#define endl '\n'
//#pragma GCC optimize ("O3")
//#pragma GCC target ("sse4")
#define SZ(x) ((int)x.size())
#define ALL(V) V.begin(), V.end()
#define L_B lower_bound
#define U_B upper_bound
#define pb push_back
using namespace std;
template<class T, class T2> inline int chkmax(T &x, const T2 &y) { return x < y ? x = y, 1 : 0; }
template<class T, class T2> inline int chkmin(T &x, const T2 &y) { return x > y ? x = y, 1 : 0; }
const int MAXN = (1 << 20);
const int B = (int)1e7 + 42;
const int inf = B + 42;
int n, q;
int dp[B + 42], cnt[inf + 42];
int a[MAXN], lp[B + 42];
void read()
{
cin >> n >> q;
for(int i = 0; i < n; i++)
cin >> a[i];
}
priority_queue<int, vector<int>, greater<int> > Q;
int last[MAXN];
void fix()
{
while(!Q.empty() && cnt[Q.top()])
{
cnt[Q.top()]--;
Q.pop();
}
}
void solve()
{
for(int i = 0; i < a[n - 1]; i++) dp[i] = 1;
for(int i = a[n - 1]; i <= B; i++) dp[i] = inf;
memset(lp, -1, sizeof(lp));
for(int i = 0; i < n; i++)
{
for(int j = 0; j <= B; j += a[i])
if(dp[j] != 1 && lp[j] == -1)
lp[j] = i;
}
for(int i = 0; i < n - 1; i++)
Q.push(1), last[i] = 1;
forward_list<int> li;
for(int d = a[n - 1]; d <= B; d++)
{
li.clear();
int dummy = d;
while(lp[dummy] != -1)
{
int D = a[lp[dummy]];
li.push_front(lp[dummy]);
while(dummy % D == 0) dummy /= D;
}
for(int i: li) cnt[last[i]]++;
fix();
if(!Q.empty())
chkmin(dp[d], Q.top() + 1);
for(int i: li)
{
Q.push(dp[d]);
last[i] = dp[d];
}
}
while(q--)
{
int x;
cin >> x;
if(dp[x] == inf) cout << "oo" << endl;
else cout << dp[x] << endl;
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
read();
solve();
return 0;
}
