Problem: A running competition is going to be held soon. The stadium where the competition will be held can be represented by several segments on the coordinate plane:

  * two horizontal segments: one connecting the points (0, 0) and (x, 0), the other connecting the points (0, y) and (x, y); 
  * n + 1 vertical segments, numbered from 0 to n. The i-th segment connects the points (a_i, 0) and (a_i, y); 0 = a_0 < a_1 < a_2 < ... < a_{n - 1} < a_n = x. 



For example, here is a picture of the stadium with x = 10, y = 5, n = 3 and a = [0, 3, 5, 10]:

<image>

A lap is a route that goes along the segments, starts and finishes at the same point, and never intersects itself (the only two points of a lap that coincide are its starting point and ending point). The length of a lap is a total distance travelled around it. For example, the red route in the picture representing the stadium is a lap of length 24.

The competition will be held in q stages. The i-th stage has length l_i, and the organizers want to choose a lap for each stage such that the length of the lap is a divisor of l_i. The organizers don't want to choose short laps for the stages, so for each stage, they want to find the maximum possible length of a suitable lap.

Help the organizers to calculate the maximum possible lengths of the laps for the stages! In other words, for every l_i, find the maximum possible integer L such that l_i mod L = 0, and there exists a lap of length exactly L.

If it is impossible to choose such a lap then print -1.

Input

The first line contains three integers n, x and y (1 ≤ n, x, y ≤ 2 ⋅ 10^5, n ≤ x).

The second line contains n + 1 integers a_0, a_1, ..., a_n (0 = a_0 < a_1 < a_2 < ... < a_{n - 1} < a_n = x).

The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of stages.

The fourth line contains q even integers l_1, l_2, ..., l_q (4 ≤ l_i ≤ 10^6) — the lengths of the stages. 

Output

Print q numbers. The i-th number should be equal to the maximum possible length of a suitable lap for the i-th stage, or -1 if it is impossible to choose a lap for that stage.

Example

Input


3 10 5
0 3 5 10
6
24 30 14 16 18 10


Output


24 30 14 16 -1 -1 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MM = 1000013;
const int P = 998244353;
const int GEN = 31;
vector<int> Div[MM];
inline int fastpow(long long x, int n) {
  long long w = 1;
  for (; n; n >>= 1, x = (x * x) % P)
    if (n & 1) w = (w * x) % P;
  return w;
}
inline int inv(int x) { return fastpow(x, P - 2); }
void FFT(vector<int> &W, long long root) {
  int N = W.size();
  if (N == 1) return;
  N >>= 1;
  vector<int> W1(N), W0(N);
  for (int i = (0); i <= ((N << 1) - 1); ++i)
    if (i & 1)
      W1[i >> 1] = W[i];
    else
      W0[i >> 1] = W[i];
  FFT(W0, root * root % P);
  FFT(W1, root * root % P);
  long long mg = 1;
  for (int i = (0); i <= (N - 1); ++i) {
    W[i] = (W0[i] + mg * W1[i]) % P;
    W[i + N] = (W0[i] + (P - mg) * W1[i]) % P;
    mg = mg * root % P;
  }
}
vector<int> mul(vector<int> A, vector<int> B) {
  int n = A.size() + B.size();
  int N = 1, ord = 0;
  while (N < n) N <<= 1, ord++;
  while (A.size() < N) A.push_back(0);
  while (B.size() < N) B.push_back(0);
  int root = fastpow(GEN, 1 << (23 - ord));
  FFT(A, root);
  FFT(B, root);
  for (int i = (0); i <= (N - 1); ++i) A[i] = 1LL * A[i] * B[i] % P;
  FFT(A, inv(root));
  int I = inv(N);
  for (int &v : A) v = 1LL * v * I % P;
  return A;
}
int T[MM], ans[MM];
vector<int> A, B, W, que;
int main() {
  ios_base::sync_with_stdio(0);
  int n, x, y, a, q;
  cin >> n >> x >> y;
  for (int i = (0); i <= (n); ++i) cin >> T[i];
  cin >> q;
  for (int i = (1); i <= (q); ++i) {
    cin >> a;
    a >>= 1;
    que.push_back(a);
  }
  for (int i = (1); i <= (500013); ++i)
    for (int j = i; j <= 500013; j += i) Div[j].push_back(i);
  A.resize(x + 1);
  B.resize(x + 1);
  for (int i = (0); i <= (n); ++i) A[T[i]] = 1, B[x - T[i]] = 1;
  W = mul(A, B);
  for (int i = (1); i <= (q); ++i) {
    a = que[i - 1];
    for (const int &v : Div[a])
      if (v > y && v - y + x < W.size() && W[v - y + x] != 0)
        ans[i] = max(ans[i], v << 1);
  }
  for (int i = (1); i <= (q); ++i) cout << (ans[i] == 0 ? -1 : ans[i]) << " ";
  return 0;
}