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:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int d, m, g[21], n, b[30001];
long long l, r;
struct info {
  std::bitset<30001> pre, suf;
  long long cnt;
  long long len;
  info() {
    pre.reset();
    suf.reset();
    cnt = len = 0;
  }
} f[101][61];
std::bitset<30001> all;
info merge(info a, info b) {
  if (!a.len) return b;
  if (!b.len) return a;
  info c;
  int l = std::max(n - b.len, 1ll), r = std::min(a.len, 1ll * (n - 1));
  c.cnt = a.cnt + b.cnt;
  if (l <= r)
    c.cnt +=
        (((a.suf & b.pre) >> l).count() - ((a.suf & b.pre) >> (r + 1)).count());
  c.len = a.len + b.len;
  c.pre = a.pre;
  if (a.len < n) c.pre &= (b.pre >> a.len) | (all << (n - a.len));
  c.suf = b.suf;
  if (b.len < n) c.suf &= (a.suf << b.len) | (all >> (30000 - b.len));
  return c;
}
info calc(long long len, int pre) {
  if (!len) return info();
  long long x = 1, p = 0;
  while (x <= len) x *= d, ++p;
  x /= d, --p;
  info ans;
  for (; len;) {
    int t = -1;
    for (long long i = 1; i * x <= len; i++) {
      ans = merge(ans, f[p][(pre + g[i]) % m]);
      t = i;
    }
    (pre += g[t + 1]) %= m;
    len %= x;
    x /= d, --p;
  }
  return ans;
}
int main() {
  scanf("%d%d", &d, &m);
  for (int i = 1; i <= d; i++) scanf("%d", g + i);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", b + i);
  scanf("%lld%lld", &l, &r);
  all.set();
  for (int i = 0; i < m; i++) {
    f[0][i].len = 1;
    f[0][i].cnt = (n == 1 && i <= b[1]);
    for (int j = 1; j < n; j++) {
      if (i <= b[n - j + 1] && j < n) f[0][i].pre.set(n - j);
      if (i <= b[j] && j) f[0][i].suf.set(j);
    }
  }
  for (long long i = 1, Mul = 1; Mul <= r; Mul *= d, i++) {
    for (int j = 0; j < m; j++) {
      f[i][j] = f[i - 1][j];
      for (int k = 2; k <= d; k++)
        f[i][j] = merge(f[i][j], f[i - 1][(j + g[k]) % m]);
    }
  }
  printf("%lld\n", calc(r, 0).cnt - calc(l + n - 2, 0).cnt);
}