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;
const int N = 30000 + 2;
bitset<N> get(int l, int r) {
  bitset<N> all1;
  if (l > r) return all1;
  all1.set();
  return (all1 >> (N - 1 - r)) ^ (all1 >> (N - 1 - (l - 1)));
}
int n, b[N];
struct Data {
  long long len, ans;
  bitset<N> pre, suf;
};
Data init(int v) {
  Data x;
  x.len = 1;
  x.ans = (n == 1) && (b[n] >= v);
  for (int i = 1; i <= n; ++i) x.pre[i] = (b[n - i + 1] >= v);
  for (int i = 1; i <= n; ++i) x.suf[i] = (b[n - i] >= v);
  return x;
}
Data operator+(const Data &l, const Data &r) {
  Data ans;
  ans.len = l.len + r.len;
  ans.ans =
      l.ans + r.ans +
      (l.suf & r.pre & get(max(1LL, n - l.len), min(r.len, n - 1LL))).count();
  if (l.len >= n)
    ans.pre = l.pre;
  else
    ans.pre = l.pre & (get(1, l.len) | (r.pre << l.len));
  if (r.len >= n)
    ans.suf = r.suf;
  else
    ans.suf = r.suf & (get(n - r.len, n) | (l.suf >> r.len));
  return ans;
}
const int D = 20 + 2, M = 60 + 2, L = 60 + 2;
int d, m, gen[D];
Data dp[L][M];
long long F(long long r) {
  Data ans;
  ans.len = ans.ans = 0;
  ans.pre.set();
  ans.suf.set();
  int v0 = 0;
  while (r) {
    int i = 0;
    while (dp[i][0].len <= r / d) ++i;
    int j = 0;
    while (r >= dp[i][0].len) {
      r -= dp[i][0].len;
      ans = ans + dp[i][(v0 + gen[j]) % m];
      ++j;
    }
    (v0 += gen[j]) %= m;
  }
  return ans.ans;
}
int main() {
  cin >> d >> m;
  for (int i = 0; i <= d - 1; ++i) scanf("%d", gen + i);
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%d", b + i);
  long long l, r;
  cin >> l >> r;
  for (int v = 0; v <= m - 1; ++v) dp[0][v] = init(v);
  int i = 0;
  while (dp[i][0].len <= r / d) {
    ++i;
    for (int v = 0; v <= m - 1; ++v) {
      dp[i][v] = dp[i - 1][v];
      for (int j = 1; j <= d - 1; ++j)
        dp[i][v] = dp[i][v] + dp[i - 1][(v + gen[j]) % m];
    }
  }
  cout << F(r) - F(l + n - 2);
}