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 MAX = 30000;
struct info {
  bitset<MAX> prefix;
  bitset<MAX> suffix;
  long long ans = 0;
  long long len = 0;
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int d, m;
  cin >> d >> m;
  vector<int> gen(d);
  for (int i = 0; i < d; ++i) {
    cin >> gen[i];
  }
  int n;
  cin >> n;
  vector<int> b(n);
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
  }
  long long l, r;
  cin >> l >> r;
  bitset<MAX> one;
  for (int i = 0; i < n - 1; ++i) {
    one[i] = 1;
  }
  auto merge = [&](const info &l, const info &r) {
    info res;
    res.prefix = l.prefix;
    res.suffix = r.suffix;
    res.ans = l.ans + r.ans;
    res.len = l.len + r.len;
    if (l.len < n - 1) {
      res.prefix &= r.prefix >> l.len | one << n - 1 - l.len;
    }
    if (r.len < n - 1) {
      res.suffix &= l.suffix << r.len | one >> n - 1 - r.len;
    }
    if (l.len + r.len >= n) {
      bitset<MAX> b = l.suffix & r.prefix;
      if (l.len < n - 1) {
        b &= one >> n - 1 - l.len;
      }
      if (r.len < n - 1) {
        b &= one << n - 1 - r.len;
      }
      res.ans += b.count();
    }
    return res;
  };
  vector<vector<info>> dp(1, vector<info>(m));
  for (int i = 0; i < m; ++i) {
    dp[0][i].len = 1;
    if (n == 1) {
      dp[0][i].ans = b[0] >= i;
    } else {
      for (int j = 0; j < n - 1; ++j) {
        if (b[j + 1] >= i) {
          dp[0][i].prefix[j] = true;
        }
        if (b[j] >= i) {
          dp[0][i].suffix[j] = true;
        }
      }
    }
  }
  long long sz = 1;
  while (sz <= r / d) {
    int cur = dp.size();
    dp.push_back(dp.back());
    for (int i = 0; i < m; ++i) {
      for (int j = 1; j < d; ++j) {
        dp[cur][i] = merge(dp[cur][i], dp[cur - 1][(i + gen[j]) % m]);
      }
    }
    sz *= d;
  }
  auto solve = [&](long long k) {
    long long cur = sz;
    int diff = 0;
    info res;
    for (int i = dp.size() - 1; ~i; --i, cur /= d) {
      for (int j = 0; j < d; ++j) {
        if (k >= cur) {
          res = merge(res, dp[i][(diff + gen[j]) % m]);
          k -= cur;
        } else {
          diff = (diff + gen[j]) % m;
          break;
        }
      }
    }
    return res.ans;
  };
  cout << solve(r) - solve(l + n - 2) << "\n";
  return 0;
}