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 ll = long long;
const int N = 30005;
int d, m, n, lm, b[N], gen[25];
ll l, r;
struct Node {
  bitset<N> p, s, a;
  ll ans, l;
  void set(int x) {
    if (x <= b[n]) p[n] = 1;
    if (x <= b[1]) s[1] = 1;
    for (int i = 1; i <= n; i++)
      if (x <= b[i]) a[i] = 1;
    ans = (n == 1 && x <= b[1]);
    l = 1;
  }
} a[65][65];
Node merge(const Node &a, const Node &b) {
  static Node ret;
  ret.l = a.l + b.l;
  ret.p = a.p;
  if (a.l < n) ret.p |= (a.a << a.l & b.p) >> a.l;
  ret.s = b.s;
  if (b.l < n) ret.s |= (b.a >> 1 & a.s) << b.l;
  if (a.l + b.l <= n)
    ret.a = b.a >> a.l & a.a;
  else
    ret.a.reset();
  ret.ans = a.ans + b.ans + (a.s << 1 & b.p).count();
  return ret;
}
ll calc(ll x) {
  ll y = 0, p = 0;
  static Node ret;
  bool fl = 0;
  ret.ans = 0;
  for (int i = lm - 1; i >= 0; --i) {
    int j;
    for (j = 0; y + a[i][(p + gen[j]) % m].l <= x; j++) {
      if (!fl)
        ret = a[i][(p + gen[j]) % m], fl = 1;
      else
        ret = merge(ret, a[i][(p + gen[j]) % m]);
      y += a[i][(p + gen[j]) % m].l;
    }
    p = (p + gen[j]) % m;
  }
  return ret.ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> d >> m;
  for (int i = 0; i < d; i++) cin >> gen[i];
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> b[i];
  cin >> l >> r;
  for (int i = 0; i < m; i++) a[0][i].set(i);
  for (lm = 1; a[lm - 1][0].l <= r / d; lm++)
    for (int i = 0; i < m; i++) {
      a[lm][i] = a[lm - 1][(i + gen[0]) % m];
      for (int j = 1; j < d; j++)
        a[lm][i] = merge(a[lm][i], a[lm - 1][(i + gen[j]) % m]);
    }
  cout << calc(r) - calc(l + n - 2) << endl;
  return 0;
}