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 maxn = 30010;
int d, M, n, tot, gen[21], B[maxn];
long long pw[maxn], L, R;
struct node {
  long long len, ans;
  bitset<maxn> pre, suf;
} f[63][63];
bitset<maxn> all;
node operator+(const node &A, const node &B) {
  if (!A.len) return B;
  node C;
  C.len = A.len + B.len;
  C.ans = A.ans + B.ans, C.pre = A.pre, C.suf = B.suf;
  if (A.len < n - 1) C.pre &= (B.pre >> A.len) | (all << (n - 1 - A.len));
  if (B.len < n - 1) C.suf &= (A.suf << B.len) | (all >> (n - 1 - B.len));
  if (A.len + B.len >= n) {
    auto b = A.suf & B.pre;
    if (A.len < n - 1) b &= all >> (n - 1 - A.len);
    if (B.len < n - 1) b &= all << (n - 1 - B.len);
    C.ans += b.count();
  }
  return C;
}
int main() {
  scanf("%d %d", &d, &M);
  for (int i = 1; i <= d; i++) {
    scanf("%d", &gen[i]);
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &B[i]);
  }
  scanf("%lld %lld", &L, &R);
  for (int i = pw[0] = 1; i < n; i++) {
    all.set(i);
  }
  for (int i = 0; i < M; i++) {
    f[0][i].len = 1, f[0][i].ans = n == 1 && B[1] >= i;
    for (int j = 1; j < n; j++) {
      f[0][i].pre[j] = B[j + 1] >= i;
      f[0][i].suf[j] = B[j] >= i;
    }
  }
  while (pw[tot] <= R / d) {
    pw[tot + 1] = pw[tot] * d, tot++;
    for (int i = 0; i < M; i++) {
      for (int j = 1; j <= d; j++) {
        f[tot][i] = f[tot][i] + f[tot - 1][(i + gen[j]) % M];
      }
    }
  }
  auto solve = [&](long long x) {
    node A;
    A.len = A.ans = 0;
    for (int i = tot, inc = 0; ~i; i--) {
      for (int j = 1; j <= d; j++) {
        if (x >= pw[i])
          x -= pw[i], A = A + f[i][(inc + gen[j]) % M];
        else {
          (inc += gen[j]) %= M;
          break;
        }
      }
    }
    return A.ans;
  };
  printf("%lld\n", solve(R) - solve(L + n - 2));
  return 0;
}