// header file
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// pragma
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
// macros
#define endl "\n"
#define ll long long
#define mp make_pair
#define ins insert
#define lb lower_bound
#define pb push_back
#define ub upper_bound
#define lll __int128
#define fi first
#define se second
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset;
typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;
int main() {
ios_base::sync_with_stdio(0); cin.tie(NULL);
int n, l;
cin >> n >> l;
int a[n + 1];
for(int i = 1; i <= n; ++i)
cin >> a[i];
int que;
cin >> que;
pair<int, int> q[que];
for(int i = 0; i < que; ++i)
cin >> q[i].fi, q[i].se = i;
sort(q, q + que);
int allow[l + 1];
memset(allow, -1, sizeof(allow));
for(int i = 0; i < que; ++i) {
if(allow[q[i].fi] == -1)
allow[q[i].fi] = i;
}
for(int i = l - 1; i >= 0; --i) {
if(allow[i] == -1)
allow[i] = allow[i + 1];
}
// given compare substring shift by x
// size l - 1
int cnt[que][n + 1];
memset(cnt, 0, sizeof(cnt));
for(int i = 1; i <= n; ++i) {
// try for every length i
// geser
int idx1 = 1, idx2 = 1 + i, diff = 0;
if(idx2 + l - 1 > n)
break;
for(int j = 0; j < l; ++j) {
if(a[idx1 + j] != a[idx2 + j])
++diff;
}
// simpan batas di query i
// find lowest yg bs ditambahkan
++cnt[allow[diff]][idx1];
++cnt[allow[diff]][idx2];
//cout << idx1 << " " << idx2 << " " << diff << endl;
for(;idx2 + l <= n;) {
// coba geser
if(a[idx1 + l] != a[idx2 + l])
++diff;
if(a[idx1] != a[idx2])
--diff;
++idx1, ++idx2;
++cnt[allow[diff]][idx1];
++cnt[allow[diff]][idx2];
//cout << idx1 << " " << idx2 << " " << diff << endl;
}
}
for(int i = 1; i < que; ++i) {
for(int j = 1; j <= n; ++j)
cnt[i][j] += cnt[i - 1][j];
}
int ans[que][n + 1];
for(int i = 0; i < que; ++i) {
for(int j = 1; j <= n; ++j) {
ans[q[i].se][j] = cnt[i][j];
}
}
for(int i = 0; i < que; ++i) {
for(int j = 1; j <= n - l + 1; ++j)
cout << ans[i][j] << " ";
cout << endl;
}
return 0;
}
