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;
const int MAX = 30010;
struct State {
  bitset<MAX> left;
  bitset<MAX> right;
  long long len = 0, res = 0;
};
int n;
bitset<MAX> all;
State Unite(const State &a, const State &b) {
  State c;
  c.len = a.len + b.len;
  c.res = a.res + b.res;
  c.left = a.left;
  c.right = b.right;
  if (a.len < n - 1) c.left &= (b.left >> a.len) | (all << (n - 1 - a.len));
  if (b.len < n - 1) c.right &= (a.right << b.len) | (all >> (n - 1 - b.len));
  if (a.len + b.len >= n) {
    bitset<MAX> d = a.right & b.left;
    if (a.len < n - 1) d &= (all >> (n - 1 - a.len));
    if (b.len < n - 1) d &= (all << (n - 1 - b.len));
    c.res += d.count();
  }
  return c;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int d, m;
  cin >> d >> m;
  vector<int> gen(d);
  for (int i = 0; i < d; i++) cin >> gen[i];
  cin >> n;
  vector<int> b(n);
  for (int i = 0; i < n; i++) cin >> b[i];
  long long L, R;
  cin >> L >> R;
  for (int i = 1; i < n; i++) all.set(i);
  vector<vector<State>> states(1, vector<State>(m));
  for (int i = 0; i < m; i++) {
    states[0][i].len = 1;
    if (n == 1)
      states[0][i].res = b[0] >= i;
    else {
      for (int j = 1; j < n; j++)
        if (b[j] >= i) states[0][i].left.set(j);
      for (int j = 1; j < n; j++)
        if (b[j - 1] >= i) states[0][i].right.set(j);
    }
  }
  long long z = 1;
  while (z <= R / d) {
    int sz = (int)states.size();
    states.resize(sz + 1);
    states[sz].resize(m);
    for (int i = 0; i < m; i++) {
      State cur = states[sz - 1][i];
      for (int j = 1; j < d; j++)
        cur = Unite(cur, states[sz - 1][(i + gen[j]) % m]);
      states[sz][i] = cur;
    }
    z *= d;
  }
  auto calc = [&](long long up) -> long long {
    State res;
    int shift = 0;
    for (int i = (int)states.size() - 1; i >= 0; i--) {
      z = 1;
      for (int j = 0; j < i; j++) z *= d;
      for (int j = 0; j < d; j++)
        if (up >= z) {
          res = Unite(res, states[i][(shift + gen[j]) % m]);
          up -= z;
        } else {
          shift = (shift + gen[j]) % m;
          break;
        }
    }
    return res.res;
  };
  cout << calc(R) - calc(L + n - 2) << '\n';
  return 0;
}