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;
const int N = 30007;
int d, m, n, g[N], w[N], st[N];
long long l, r;
struct node {
  bitset<30007> a, b;
  long long len, sum;
} f[100][100];
bitset<30007> all, opt;
inline node merge(node a, node b) {
  node c;
  c.len = a.len + b.len, c.sum = a.sum + b.sum;
  c.a = (a.a & ((all >> min(a.len, 30000ll)) ^ all)) |
        ((b.a >> min(a.len, 30000ll)) & a.a);
  c.b = (b.b & ((all << min(b.len, 30000ll)) ^ all)) |
        ((a.b << min(b.len, 30000ll)) & b.b);
  opt = (a.b & (b.a >> 1));
  opt = (opt & ((all << min(a.len, 30000ll)) ^ all));
  opt = (opt & ((all >> min(b.len + 1, 30001ll)) ^ all));
  c.sum = c.sum + opt.count();
  return c;
}
inline long long solve(long long len) {
  int c = 0, k = 1;
  st[0] = 0;
  while (len > 0) st[++st[0]] = len % d, len /= d;
  for (int i = 1; i <= st[0]; i++) {
    if (i > 1) {
      for (int j = 0; j < m; j++) {
        f[i][j] = f[i - 1][j];
        for (int x = 1; x < d; x++) {
          f[i][j] = merge(f[i][j], f[i - 1][(j + g[x]) % m]);
        }
      }
    } else {
      for (int j = 0; j < m; j++) {
        f[i][j] = f[i - 1][j];
        for (int x = 1; x < d; x++) {
          f[i][j] = merge(f[i][j], f[i - 1][(j + g[x]) % m]);
        }
      }
    }
  }
  int opt = 0;
  bool fl = 0;
  node ans;
  for (int i = st[0]; i >= 1; i--) {
    if (i > 1) {
      for (int j = 0; j < st[i]; j++) {
        if (!fl)
          ans = f[i - 1][(g[j] + opt) % m], fl = 1;
        else
          ans = merge(ans, f[i - 1][(g[j] + opt) % m]);
      }
    } else {
      for (int j = 0; j < st[i]; j++) {
        if (!fl)
          ans = f[0][(g[j] + opt) % m], fl = 1;
        else
          ans = merge(ans, f[i - 1][(g[j] + opt) % m]);
      }
    }
    opt += g[st[i]];
  }
  if (!fl) return 0;
  return ans.sum;
}
int main() {
  cin >> d >> m;
  for (int i = 0; i < d; i++) scanf("%d", &g[i]);
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]), all[i] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= w[i]; j++) {
      f[0][j].a[i] = f[0][j].b[i] = 1;
      if (n == 1) f[0][j].sum++;
    }
  for (int i = 0; i < m; i++) f[0][i].len = 1;
  cin >> l >> r;
  cout << solve(r) - solve(l + n - 2ll) << endl;
}
