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 oo = 0x3f3f3f3f;
const long long ooo = 9223372036854775807ll;
const int _cnt = 1000 * 1000 + 7;
const int _p = 1000 * 1000 * 1000 + 7;
const int N = 30005;
const double PI = acos(-1.0);
const double eps = 1e-9;
int o(int x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
void file_put() {
  string s = "/home/jslijin/jslijin/code/";
  freopen((s + "code.in").c_str(), "r", stdin);
  freopen((s + "code.out").c_str(), "w", stdout);
}
int n, m, d, B[N], c = 0, gen[25], delta;
long long P[65], l, r;
bitset<N> one, tmp;
struct node {
  long long num, len;
  bitset<N> pre, suf;
  node() {
    num = len = 0;
    pre = suf = 0;
  }
} f[65][60];
inline node operator+(node &a, node &b) {
  if (!a.len) return b;
  node ret;
  ret.len = a.len + b.len;
  ret.num = a.num + b.num;
  ret.pre = a.pre, ret.suf = b.suf;
  if (a.len < n - 1) ret.pre &= (b.pre >> a.len) | (one << n - 1 - a.len);
  if (b.len < n - 1) ret.suf &= (a.suf << b.len) | (one >> n - 1 - b.len);
  if (ret.len < n) return ret;
  tmp = a.suf & b.pre;
  if (a.len < n - 1) tmp &= one >> n - 1 - a.len;
  if (b.len < n - 1) tmp &= one << n - 1 - b.len;
  ret.num += tmp.count();
  return ret;
}
long long solve(long long n) {
  for (c = 0, delta = 0; P[c + 1] <= n; ++c)
    ;
  node ret;
  for (int i = (c); i >= (0); --i)
    for (int k = (0); k <= (d - 1); ++k)
      if (n >= P[i]) {
        ret = ret + f[i][(delta + gen[k]) % m];
        n -= P[i];
        if (!n) return ret.num;
      } else {
        delta += gen[k];
        break;
      }
  return ret.num;
}
int main() {
  scanf("%d%d", &d, &m);
  for (int i = (0); i <= (d - 1); ++i) scanf("%d", &gen[i]);
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) scanf("%d", &B[i]);
  scanf("%lld%lld", &l, &r);
  P[0] = 1;
  for (int i = (1); i <= (64); ++i) P[i] = P[i - 1] * d;
  for (; P[c + 1] <= r; ++c)
    ;
  for (int i = (1); i <= (n - 1); ++i) one[i] = 1;
  for (int j = (0); j <= (m - 1); ++j) {
    f[0][j].len = 1;
    if (n == 1)
      f[0][j].num = j <= B[1];
    else
      for (int k = (1); k <= (n - 1); ++k)
        f[0][j].pre[k] = j <= B[k + 1], f[0][j].suf[k] = j <= B[k];
  }
  for (int i = (1); i <= (c); ++i)
    for (int j = (0); j <= (m - 1); ++j)
      for (int k = (0); k <= (d - 1); ++k)
        f[i][j] = f[i][j] + f[i - 1][(j + gen[k]) % m];
  printf("%lld\n", solve(r) - solve(l + n - 2));
  return 0;
}