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;
int d, m;
vector<int> gen;
vector<int> b;
map<long long, vector<long long> > dp_saved;
long long cnt(long long S, vector<long long> masks) {
  assert(masks.size() > 0);
  if (S == 0) return 0;
  if (masks.size() == 1) {
    if (!dp_saved.count(S)) {
      long long Suse = S;
      vector<long long> digits;
      while (Suse > 0) {
        digits.push_back(Suse % d);
        Suse /= d;
      }
      reverse(digits.begin(), digits.end());
      vector<long long> dp(m, 0);
      long long cursum = 0;
      for (long long digit : digits) {
        vector<long long> newdp(m, 0);
        for (int i = 0; i < m; i++) {
          for (int j = 0; j < d; j++) {
            newdp[(i + gen[j]) % m] += dp[i];
          }
        }
        for (int j = 0; j < digit; j++) {
          newdp[(cursum + gen[j]) % m]++;
        }
        dp = newdp;
        cursum += gen[digit];
      }
      dp_saved[S] = dp;
    }
    vector<long long> dp = dp_saved[S];
    long long ret = 0;
    for (int i = 0; i < m; i++) {
      if (masks[0] & (1LL << i)) {
        ret += dp[i];
      }
    }
    return ret;
  } else {
    long long ret = 0;
    for (int xres = 0; xres < d; xres++) {
      long long len = masks.size();
      long long newlen = (len - 1 + xres) / d + 1;
      vector<long long> newmasks(newlen, (1LL << m) - 1);
      for (long long i = 0; i < len; i++) {
        long long newidx = (i + xres) / d;
        long long diff = gen[(i + xres) % d];
        newmasks[newidx] &= (masks[i] >> diff) | (masks[i] << (m - diff));
      }
      long long newS = (S - xres + d - 1) / d;
      ret += cnt(newS, newmasks);
    }
    return ret;
  }
}
long long solve(long long maxs) {
  vector<long long> masks;
  for (int r : b) {
    long long mask = (1LL << m) - 1;
    for (int z = 0; z < m; z++) {
      if (z > r) {
        mask &= ~(1LL << z);
      }
    }
    masks.push_back(mask);
  }
  return cnt(maxs, masks);
}
int main() {
  cin >> d >> m;
  gen.resize(d);
  for (int i = 0; i < d; i++) {
    cin >> gen[i];
  }
  int n;
  cin >> n;
  b.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  long long l, r;
  cin >> l >> r;
  l--;
  r--;
  long long ans = solve(r - n + 2) - solve(l);
  cout << ans << endl;
}