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>
using namespace std;
int n, m, D, gen[103], B[30003];
long long pw[63];
int bit;
struct dat {
  bitset<30003> pre, suf;
  long long ans, len;
  void create(int val) {
    ans = 0, len = 1;
    if (n == 1)
      ans = (val <= B[0]);
    else {
      for (int i = 0; i < n; ++i) pre[i] = suf[i] = (val <= B[i]);
    }
  }
  friend dat operator+(dat p, dat q) {
    dat r;
    r.len = p.len + q.len;
    long long lp = max(0ll, n - p.len), lq = max(0ll, 30003 - q.len);
    r.pre = (p.pre >> lp << lp) | ((q.pre & (p.pre << p.len)) >> p.len);
    r.suf = (q.suf << lq >> lq) | ((p.suf & (q.suf >> q.len)) << q.len);
    lp = max(30003 - p.len, 0ll), lq = max(n - q.len, 0ll);
    r.ans = p.ans + q.ans +
            ((p.suf << lp >> lp) << 1 & (q.pre >> lq << lq)).count();
    return r;
  }
} f[63][60];
long long L, R;
void init() {
  pw[bit = 0] = 1;
  for (long long tms = 1; tms <= R / D; pw[++bit] = (tms *= D))
    ;
  for (int j = 0; j < m; ++j) f[0][j].create(j);
  for (int i = 1; i <= bit; ++i)
    for (int j = 0; j < m; ++j) {
      f[i][j] = f[i - 1][j];
      for (int k = 1; k < D; ++k)
        f[i][j] = f[i][j] + f[i - 1][(j + gen[k]) % m];
    }
}
dat solve(long long LIM, int x, int add) {
  if (LIM < pw[x]) return solve(LIM, x - 1, add);
  dat res = f[x][add];
  LIM -= pw[x];
  for (int i = 1; LIM && i < D; ++i) {
    if (LIM < pw[x])
      res = res + solve(LIM, x - 1, (add + gen[i]) % m), LIM = 0;
    else
      res = res + f[x][(add + gen[i]) % m], LIM -= pw[x];
  }
  return res;
}
int main() {
  cin >> D >> m;
  for (int i = 0; i < D; ++i) cin >> gen[i];
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> B[i];
  cin >> L >> R;
  init();
  cout << solve(R, bit, 0).ans - (L == 1 ? 0 : solve(L + n - 2, bit, 0).ans)
       << endl;
  return 0;
}