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 MAXN = 3e4 + 5;
const int MAXD = 25;
const int MAXM = 65;
const int MAXLOG = 64;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
struct info {
  long long ans, len;
  bitset<MAXN> pre, suf;
};
bitset<MAXN> all;
int n, d, m, gen[MAXD], b[MAXN];
int tot;
long long len[MAXLOG];
info dp[MAXLOG][MAXM];
info operator+(const info &a, const info &b) {
  if (a.len == 0) return b;
  info res;
  res.len = a.len + b.len;
  res.ans = a.ans + b.ans;
  res.pre = a.pre;
  res.suf = b.suf;
  if (a.len <= n - 2) res.pre &= (b.pre >> a.len) | (all << (n - 1 - a.len));
  if (b.len <= n - 2) res.suf &= (a.suf << b.len) | (all >> (n - 1 - b.len));
  if (a.len + b.len >= n) {
    bitset<MAXN> tmp = a.suf & b.pre;
    if (a.len <= n - 2) tmp &= all >> (n - 1 - a.len);
    if (b.len <= n - 2) tmp &= all << (n - 1 - b.len);
    res.ans += tmp.count();
  }
  return res;
}
long long work(long long x) {
  info res;
  res.ans = 0;
  res.len = 0;
  res.pre.reset();
  res.suf.reset();
  int add = 0;
  for (int i = tot; i >= 0; i--) {
    for (int j = 1; j <= d; j++)
      if (x >= len[i]) {
        x -= len[i];
        res = res + dp[i][(add + gen[j]) % m];
      } else {
        add = (add + gen[j]) % m;
        break;
      }
  }
  return res.ans;
}
int main() {
  read(d), read(m);
  for (int i = 1; i <= d; i++) read(gen[i]);
  read(n);
  for (int i = 1; i <= n; i++) read(b[i]);
  for (int i = 1; i <= n - 1; i++) all.set(i);
  tot = 0, len[0] = 1;
  for (int i = 0; i <= m - 1; i++) {
    dp[0][i].len = 1;
    if (n == 1)
      dp[0][i].ans = i <= b[1];
    else {
      for (int j = 1; j <= n - 1; j++) {
        dp[0][i].pre[j] = i <= b[j + 1];
        dp[0][i].suf[j] = i <= b[j];
      }
    }
  }
  long long ql, qr;
  read(ql), read(qr);
  while (len[tot] <= qr / d) {
    tot++, len[tot] = len[tot - 1] * d;
    for (int i = 0; i <= m - 1; i++)
      for (int j = 1; j <= d; j++)
        dp[tot][i] = dp[tot][i] + dp[tot - 1][(i + gen[j]) % m];
  }
  writeln(work(qr) - work(ql + n - 2));
  return 0;
}
