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;
int gen[20];
vector<long long int> B;
long long int dp[100][60];
long long int shift(long long int b, int x) {
  return (b >> x) | ((b & ((1LL << x) - 1)) << (m - x));
}
map<long long int, int> M;
int num = 0;
long long int ans[100][60];
int xx[100][60];
long long int solve(vector<long long int> v, long long int r, int s) {
  if (r == 0)
    return 0;
  else if (v.size() == 1) {
    int i = 0;
    if (r == 1) return (v[0] & 1) ? 1 : 0;
    if (!M.count(r)) {
      int j, k, c = 0, digits[100];
      long long int x = r - 1;
      while (x > 0) digits[i++] = x % d, x /= d;
      for (i--; i >= 0; i--) {
        for (j = 0; j < digits[i]; j++) {
          for (k = 0; k < m; k++)
            ans[num][k] += dp[i][(((k - c - gen[j]) % m) + m) % m];
        }
        if (i == 0) {
          for (k = 0; k < m; k++)
            ans[num][k] += dp[i][(((k - c - gen[j]) % m) + m) % m];
        }
        c += gen[j], c %= m;
      }
      M[r] = num++;
    }
    int p = M[r];
    for (i = 0; i < m; i++) {
      if (v[0] & (1LL << i)) xx[p][i] += s;
    }
    return 0;
  }
  int i, j;
  long long int ans = 0;
  for (i = 0; i < v.size(); i++) {
    if (v[i] == 0) return 0;
  }
  for (i = 0; i < d; i++) {
    vector<long long int> v2((i + v.size() - 1) / d + 1, (1LL << m) - 1);
    for (j = 0; j < v.size(); j++)
      v2[(j + i) / d] &= shift(v[j], gen[(i + j) % d]);
    ans += solve(v2, (r - i + d - 1) / d, s);
  }
  return ans;
}
int main() {
  int i;
  int n;
  long long int l, r;
  scanf("%d %d", &d, &m);
  for (i = 0; i < d; i++) scanf("%d", &gen[i]);
  scanf("%d", &n), B.resize(n);
  for (i = 0; i < n; i++) scanf("%I64d", &B[i]), B[i] = (1LL << (B[i] + 1)) - 1;
  scanf("%I64d %I64d", &l, &r);
  int j, k;
  dp[0][0] = 1;
  for (i = 1; i < 100; i++) {
    if (pow(d, i) > 3e18) break;
    for (j = 0; j < m; j++) {
      for (k = 0; k < d; k++) dp[i][j] += dp[i - 1][(j - gen[k] + m) % m];
    }
  }
  long long int sum = solve(B, r - n + 1, 1) - solve(B, l - 1, -1);
  for (i = 0; i < num; i++) {
    for (j = 0; j < m; j++) sum += xx[i][j] * ans[i][j];
  }
  printf("%I64d\n", sum);
  return 0;
}