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:

<image>
#include <bits/stdc++.h>
using namespace std;
int d, m, n, x[25], a[30005];
vector<long long> ok;
long long L, R;
map<long long, vector<long long> > ans;
map<long long, vector<long long> > zip;
vector<long long> calc(long long lim) {
  if (zip.find(lim) != zip.end()) return zip[lim];
  vector<long long> ret(m, 0);
  if (lim == 0) {
    ret[0] = 1;
    return ret;
  }
  for (int beg = 0; beg < d && beg <= lim; beg++) {
    long long nxt_lim = (lim - beg) / d;
    vector<long long> R = calc(nxt_lim);
    for (int i = 0; i < m; i++) {
      int a = (i + x[beg]);
      if (a >= m) a -= m;
      ret[a] += R[i];
    }
  }
  return zip[lim] = ret;
}
void rec(long long lim, vector<long long> mask, bool F) {
  if (mask.size() == 1) {
    if (ans.find(lim) == ans.end()) {
      ans[lim] = vector<long long>(m, 0);
    }
    for (int i = 0; i < m; i++) {
      if (((mask[0] >> i) & 1LL)) {
        if (F)
          ans[lim][i]++;
        else
          ans[lim][i]--;
      }
    }
    return;
  }
  for (int beg = 0; beg < d && beg <= lim; beg++) {
    long long nxt_lim = (lim - beg) / d;
    vector<long long> vec;
    int zan = mask.size();
    int bb = (d - beg);
    int xx = 1 + (zan - bb + d - 1) / d;
    vec.resize(xx, 0LL);
    long long B = (1LL << m) - 1;
    int cur = beg;
    int id = 0;
    for (int i = 0; i < mask.size(); i++) {
      int add = x[cur];
      long long nw = 0;
      nw = (mask[i] >> add) | (mask[i] << (m - add));
      B &= nw;
      if (cur == d - 1 || i == mask.size() - 1) {
        vec[id++] = B;
        B = (1LL << m) - 1;
      }
      cur++;
      if (cur == d) cur = 0;
    }
    assert(id == xx);
    rec(nxt_lim, vec, F);
  }
}
int main() {
  cin >> d >> m;
  for (int i = 0; i < d; i++) {
    cin >> x[i];
  }
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ok.push_back((2LL << a[i]) - 1LL);
  }
  cin >> L >> R;
  L--;
  R--;
  rec(R - n + 1, ok, true);
  if (L - 1 >= 0) rec(L - 1, ok, false);
  long long ret = 0;
  for (map<long long, vector<long long> >::iterator it = ans.begin();
       it != ans.end(); it++) {
    vector<long long> tbl = calc(it->first);
    for (int i = 0; i < m; i++) {
      ret += tbl[i] * it->second[i];
    }
  }
  cout << ret << endl;
}
