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 d, m, n;
int gen[20];
int b[30000];
long long l, r;
struct Info {
  long long len, ans;
  bitset<30000 + 1> p, s;
} dp[70][60];
long long pw[70];
void merge(Info &c, Info &a, Info &b) {
  if (a.len == 0) {
    c = b;
    return;
  }
  c.ans = a.ans + b.ans;
  if (n - a.len > 1 || b.len < n - 1)
    for (int i = max(n - a.len, 1ll); i <= b.len && i < n; ++i)
      c.ans += (a.s[i] & b.p[i]);
  else
    c.ans += (a.s & b.p).count();
  static bitset<30000 + 1> tmp;
  if (n - 1 - a.len > 0) {
    tmp = a.p & b.p << a.len;
    for (int i = 1; i <= a.len; ++i) tmp[i] = a.p[i];
    c.p = tmp;
  } else
    c.p = a.p;
  if (1 + b.len < n) {
    tmp = b.s & a.s >> b.len;
    for (int i = n - b.len; i < n; ++i) tmp[i] = b.s[i];
    c.s = tmp;
  } else
    c.s = b.s;
  c.len = a.len + b.len;
}
long long calc(long long r) {
  if (r <= 0) return 0;
  static Info res;
  res.len = 0;
  res.ans = 0;
  long long p = 1, w = 0;
  for (; p <= r; p *= d, ++w)
    ;
  --w;
  int v = 0;
  for (; w >= 0; --w) {
    for (int i = 0; i < r / pw[w]; ++i)
      merge(res, res, dp[w][(v + gen[i]) % m]);
    v = (v + gen[r / pw[w]]) % m;
    r %= pw[w];
  }
  return res.ans;
}
int main() {
  scanf("%d%d", &d, &m);
  for (int i = 0; i < d; ++i) scanf("%d", &gen[i]);
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &b[i]);
  scanf("%lld%lld", &l, &r);
  if (r - l + 1 < n) {
    printf("0\n");
    return 0;
  }
  for (int v = 0; v < m; ++v) {
    dp[0][v].len = 1;
    if (n == 1) dp[0][v].ans = (b[0] >= v);
    for (int i = 1; i < n; ++i)
      if (b[n - i] >= v) dp[0][v].p[i] = 1;
    for (int i = 1; i < n; ++i)
      if (b[i - 1] >= v) dp[0][v].s[n - i] = 1;
  }
  pw[0] = 1;
  long long p = d;
  for (int i = 1; p <= r; ++i, p *= d) {
    pw[i] = p;
    for (int v = 0; v < m; ++v) {
      dp[i][v] = dp[i - 1][v];
      for (int j = 1; j < d; ++j)
        merge(dp[i][v], dp[i][v], dp[i - 1][(v + gen[j]) % m]);
    }
  }
  printf("%lld\n", calc(r) - calc(l + n - 2));
  return 0;
}