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;
const int N = 30000;
template <typename T>
void read(T &x) {
  int ch = getchar();
  x = 0;
  bool f = false;
  for (; ch < '0' || ch > '9'; ch = getchar()) f |= ch == '-';
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  if (f) x = -x;
}
int d, m, n, a[20], b[N], tot;
bitset<N> all;
long long l, r;
struct Node {
  long long ans, len;
  bitset<N> pre, suf;
  Node() {
    ans = len = 0;
    pre.reset();
    suf.reset();
  }
  Node operator+(const Node &o) const {
    Node c;
    c.ans = ans + o.ans;
    c.len = len + o.len;
    c.pre = pre;
    c.suf = o.suf;
    if (len < n - 1) c.pre &= (o.pre >> len) | (all << (n - 1 - len));
    if (o.len < n - 1) c.suf &= (suf << o.len) | (all >> (n - 1 - o.len));
    if (c.len >= n) {
      bitset<N> tmp = suf & o.pre;
      if (len < n - 1) tmp &= all >> (n - 1 - len);
      if (o.len < n - 1) tmp &= all << (n - 1 - o.len);
      c.ans += tmp.count();
    }
    return c;
  }
  Node operator+=(const Node &o) { return *this = *this + o; }
} f[60][60];
long long qry(long long x) {
  Node r;
  int now = 0;
  for (int i = tot; ~i; --i)
    for (int j = 0; j < d; ++j)
      if (x >= f[i][0].len) {
        r += f[i][(now + a[j]) % m];
        x -= f[i][0].len;
      } else {
        now = (now + a[j]) % m;
        break;
      }
  return r.ans;
}
int main() {
  read(d);
  read(m);
  for (int i = 0; i < d; ++i) read(a[i]);
  read(n);
  for (int i = 0; i < n; ++i) read(b[i]);
  for (int i = 0; i < n - 1; ++i) all.set(i);
  for (int i = 0; i < m; ++i) {
    f[0][i].len = 1;
    if (n == 1)
      f[0][i].ans = i <= b[0];
    else
      for (int j = 0; j < n - 1; ++j) {
        f[0][i].pre[j] = i <= b[j + 1];
        f[0][i].suf[j] = i <= b[j];
      }
  }
  read(l);
  read(r);
  while (d * f[tot][0].len <= r) {
    ++tot;
    for (int i = 0; i < m; ++i) {
      f[tot][i] = f[tot - 1][i];
      for (int j = 1; j < d; ++j) f[tot][i] += f[tot - 1][(i + a[j]) % m];
    }
  }
  printf("%lld\n", qry(r) - qry(l + n - 2));
}