Problem: A Thue-Morse-Radecki-Mateusz sequence (Thorse-Radewoosh sequence in short) is an infinite sequence constructed from a finite sequence gen of length d and an integer m, obtained in the following sequence of steps:

  * In the beginning, we define the one-element sequence M_0=(0).
  * In the k-th step, k ≥ 1, we define the sequence M_k to be the concatenation of the d copies of M_{k-1}. However, each of them is altered slightly — in the i-th of them (1 ≤ i ≤ d), each element x is changed to (x+gen_i) \pmod{m}. 



For instance, if we pick gen = (0, \color{blue}{1}, \color{green}{2}) and m = 4: 

  * M_0 = (0), 
  * M_1 = (0, \color{blue}{1}, \color{green}{2}), 
  * M_2 = (0, 1, 2, \color{blue}{1, 2, 3}, \color{green}{2, 3, 0}), 
  * M_3 = (0, 1, 2, 1, 2, 3, 2, 3, 0, \color{blue}{1, 2, 3, 2, 3, 0, 3, 0, 1}, \color{green}{2, 3, 0, 3, 0, 1, 0, 1, 2}), and so on. 



As you can see, as long as the first element of gen is 0, each consecutive step produces a sequence whose prefix is the sequence generated in the previous step. Therefore, we can define the infinite Thorse-Radewoosh sequence M_∞ as the sequence obtained by applying the step above indefinitely. For the parameters above, M_∞ = (0, 1, 2, 1, 2, 3, 2, 3, 0, 1, 2, 3, 2, 3, 0, 3, 0, 1, ...).

Mateusz picked a sequence gen and an integer m, and used them to obtain a Thorse-Radewoosh sequence M_∞. He then picked two integers l, r, and wrote down a subsequence of this sequence A := ((M_∞)_l, (M_∞)_{l+1}, ..., (M_∞)_r).

Note that we use the 1-based indexing both for M_∞ and A.

Mateusz has his favorite sequence B with length n, and would like to see how large it is compared to A. Let's say that B majorizes sequence X of length n (let's denote it as B ≥ X) if and only if for all i ∈ \{1, 2, ..., n\}, we have B_i ≥ X_i.

He now asks himself how many integers x in the range [1, |A| - n + 1] there are such that B ≥ (A_x, A_{x+1}, A_{x+2}, ..., A_{x+n-1}). As both sequences were huge, answering the question using only his pen and paper turned out to be too time-consuming. Can you help him automate his research?

Input

The first line contains two integers d and m (2 ≤ d ≤ 20, 2 ≤ m ≤ 60) — the length of the sequence gen and an integer used to perform the modular operations. The second line contains d integers gen_i (0 ≤ gen_i < m). It's guaranteed that the first element of the sequence gen is equal to zero.

The third line contains one integer n (1 ≤ n ≤ 30000) — the length of the sequence B. The fourth line contains n integers B_i (0 ≤ B_i < m). The fifth line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}, r-l+1 ≥ n).

Output

Print a single integer — the answer to the problem.

Examples

Input


2 2
0 1
4
0 1 1 0
2 21


Output


6


Input


3 4
0 1 2
2
0 2
6 11


Output


1

Note

Thorse-Radewoosh sequence in the first example is the standard Thue-Morse sequence, so the sequence A is as follows: 11010011001011010010. Here are the places where the sequence B majorizes A:

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using cat = long long;
int D, M;
vector<int> gen;
vector<vector<vector<cat> > > cnt;
void solve_cnt(int max_dig) {
  vector<vector<cat> > cnt_sum(max_dig + 1, vector<cat>(M, 0));
  cnt_sum[0][0] = 1;
  for (int i = 0; i < max_dig; i++)
    for (int j = 0; j < M; j++)
      for (int k = 0; k < D; k++)
        cnt_sum[i + 1][(j + gen[k]) % M] += cnt_sum[i][j];
  cnt.resize(max_dig + 1, vector<vector<cat> >(D, vector<cat>(M, 0)));
  for (int i = 0; i < max_dig; i++)
    for (int j = 0; j < M; j++)
      for (int k = 0; k < D; k++) cnt[i][k][(j + gen[k]) % M] += cnt_sum[i][j];
  for (int i = 0; i <= max_dig; i++)
    for (int j = 0; j < M; j++)
      for (int k = 1; k < D; k++) cnt[i][k][j] += cnt[i][k - 1][j];
}
cat solve_leaf(cat A_mask, cat Xl, cat Xr) {
  cat ret = 0;
  static vector<char> Vl;
  Vl.clear();
  static vector<char> Vr;
  Vr.clear();
  while (Xl) {
    Vl.push_back(Xl % D);
    Xl /= D;
  }
  while (Xr) {
    Vr.push_back(Xr % D);
    Xr /= D;
  }
  while (!Vl.empty() && !Vr.empty() && Vl.size() == Vr.size() &&
         Vl.back() == Vr.back()) {
    int s = gen[Vl.back()];
    A_mask = (A_mask >> s) | ((A_mask & ((1LL << s) - 1)) << (M - s));
    Vl.pop_back();
    Vr.pop_back();
  }
  for (int i = (int)Vr.size() - 1, s = 0; i >= 0; i--) {
    cat mask = (A_mask >> s) | ((A_mask & ((1LL << s) - 1)) << (M - s));
    if (Vr[i])
      for (int k = 0; mask; k++, mask >>= 1)
        ret += cnt[i][Vr[i] - 1][k] * (mask & 1);
    s += gen[Vr[i]];
    if (s >= M) s -= M;
  }
  for (int i = (int)Vl.size() - 1, s = 0; i >= 0; i--) {
    cat mask = (A_mask >> s) | ((A_mask & ((1LL << s) - 1)) << (M - s));
    if (Vl[i])
      for (int k = 0; mask; k++, mask >>= 1)
        ret -= cnt[i][Vl[i] - 1][k] * (mask & 1);
    s += gen[Vl[i]];
    if (s >= M) s -= M;
  }
  return ret;
}
cat solve_recursive(vector<cat>& A_mask, cat Xl, cat Xr) {
  int N = A_mask.size();
  cat ret = 0;
  for (int i = 0; i < D; i++) {
    if (Xr - 1 - i < 0) break;
    vector<cat> A_mask_new;
    A_mask_new.reserve(A_mask.size() / D + 2);
    for (int j = 0, k = i, l = 0; j < N; j++, k++) {
      if (k == D) k = 0, l++;
      int g = gen[k];
      cat mask_new =
          (A_mask[j] >> g) | ((A_mask[j] & ((1LL << g) - 1)) << (M - g));
      if ((int)A_mask_new.size() == l) A_mask_new.push_back((1LL << M) - 1);
      A_mask_new[l] &= mask_new;
    }
    if (A_mask_new.size() == 1)
      ret += solve_leaf(A_mask_new[0],
                        (Xl - 1 - i >= 0) ? (Xl - 1 - i) / D + 1 : 0,
                        (Xr - 1 - i) / D + 1);
    else
      ret += solve_recursive(A_mask_new,
                             (Xl - 1 - i >= 0) ? (Xl - 1 - i) / D + 1 : 0,
                             (Xr - 1 - i) / D + 1);
  }
  return ret;
}
cat solve(vector<int>& B, cat Xl, cat Xr) {
  int N = B.size();
  vector<cat> A_mask(N);
  for (int i = 0; i < N; i++) A_mask[i] = (1LL << (B[i] + 1)) - 1;
  if (N == 1)
    return solve_leaf(A_mask[0], Xl, Xr);
  else
    return solve_recursive(A_mask, Xl, Xr);
}
int main() {
  cin.sync_with_stdio(0);
  cin >> D >> M;
  gen.resize(D);
  for (int i = 0; i < D; i++) cin >> gen[i];
  solve_cnt(100);
  int N;
  cin >> N;
  vector<int> B(N);
  for (int i = 0; i < N; i++) cin >> B[i];
  cat L, R;
  cin >> L >> R;
  L--, R--;
  cout << solve(B, L, R - N + 2) << "\n";
}