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;
using ll = long long;
const int MD = 22;
const int MM = 65;
const int MN = 3e4 + 100;
const ll INF = 1e18;
using vb = bitset<MN>;
vb F[MN];
int D, M, N;
int g[MD], b[MN];
int md(int x) {
  if (x >= M) x -= M;
  assert(x < M);
  return x;
}
struct DAT {
  vb p, s;
  ll l, c;
  DAT() {
    p.reset(), s.reset();
    l = c = 0LL;
  }
  DAT operator+(const DAT& o) const {
    DAT r;
    r.c = c + o.c, r.l = l + o.l;
    if (l + 1 < N)
      r.p = p & o.p >> l | p >> N - l - 1 << N - l - 1;
    else
      r.p = p;
    if (o.l + 1 < N)
      r.s = o.s & s << o.l | o.s & F[o.l];
    else
      r.s = o.s;
    if (r.l >= N) {
      vb q = s & o.p;
      if (l < N) q &= F[l];
      if (o.l < N) q >>= N - o.l - 1;
      r.c += q.count();
    }
    return r;
  }
};
DAT d[100][MM];
using vi = vector<int>;
ll solve(ll v) {
  if (v <= 0) return 0LL;
  vi x(0);
  for (; v; v /= D) x.push_back(v % D);
  DAT c;
  for (int i = x.size() - 1, r = 0; i >= 0; i--) {
    for (int j = 0; j < x[i]; j++) c = c + d[i][md(r + g[j])];
    r = md(r + g[x[i]]);
  }
  return c.c;
}
int main() {
  F[0].reset();
  for (int i = 0; i < MN - 1; i++) F[i + 1] = F[i], F[i + 1][i] = 1;
  scanf("%d%d", &D, &M);
  for (int i = 0; i < D; i++) scanf("%d", g + i);
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", b + i);
  b[N] = -1;
  for (int i = 0; i < M; i++) {
    d[0][i].l = 1;
    if (N > 1)
      for (int j = 0; j < N - 1; j++)
        d[0][i].s[j] = i <= b[j], d[0][i].p[j] = i <= b[j + 1];
    else
      d[0][i].c = i <= b[0];
  }
  ll x = 1;
  for (int i = 1; x <= INF / D; i++, x *= D)
    for (int j = 0; j < M; j++) {
      d[i][j] = d[i - 1][j];
      for (int k = 1; k < D; k++) d[i][j] = d[i][j] + d[i - 1][md(j + g[k])];
    }
  ll A, B;
  scanf("%lld%lld", &A, &B);
  printf("%lld\n", solve(B) - solve(A + N - 2));
  return 0;
}