#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
template<class T, int SZ> struct Seg {
T seg[2*SZ], MN = 0;
Seg() {
F0R(i,2*SZ) seg[i] = MOD;
}
T comb(T a, T b) { return min(a,b); } // easily change this to min or max
void upd(int p, T value) { // set value at position p
for (seg[p += SZ] = value; p > 1; p >>= 1)
seg[p>>1] = comb(seg[(p|1)^1],seg[p|1]); // non-commutative operations
}
void build() {
F0Rd(i,SZ) seg[i] = comb(seg[2*i],seg[2*i+1]);
}
T query(int l, int r) { // sum on interval [l, r]
T res1 = MN, res2 = MN; r++;
for (l += SZ, r += SZ; l < r; l >>= 1, r >>= 1) {
if (l&1) res1 = comb(res1,seg[l++]);
if (r&1) res2 = comb(seg[--r],res2);
}
return comb(res1,res2);
}
};
Seg<int,1<<17> S;
int m,Q, mx, ans[10000001], p[MX];
priority_queue<pi,vpi,greater<pi>> use;
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> m >> Q;
F0R(i,m) {
cin >> p[i];
use.push({p[i],i});
S.upd(i,0);
}
vi todo;
FOR(i,1,10000001) {
while (sz(use) && use.top().f == i) {
S.upd(use.top().s,MOD);
todo.pb(use.top().s);
use.pop();
}
if (sz(use) == 0) {
mx = i-1;
break;
}
ans[i] = S.seg[1]+1;
for (int j: todo) {
S.upd(j,ans[i]);
use.push({p[j]+i,j});
}
todo.clear();
}
F0R(i,Q) {
int x; cin >> x;
if (x > mx) cout << "oo";
else cout << ans[x];
cout << "\n";
}
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
