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>
const int N = 3e4 + 1;
using namespace std;
long long ri() {
  char c = getchar();
  long long x = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) - '0' + c;
  return x * f;
}
long long len[66];
int n, d, m, tot, gen[25], b[N];
std::bitset<N> all;
struct ooxx {
  long long ans, len;
  bitset<30001> pre, nxt;
  ooxx operator+(const ooxx &emm) {
    if (!len) return emm;
    ooxx ret;
    ret.ans = ans + emm.ans;
    ret.len = len + emm.len;
    ret.pre = pre;
    ret.nxt = emm.nxt;
    if (len < n - 1) ret.pre &= ((emm.pre >> len) | (all << (n - len - 1)));
    if (emm.len < n - 1)
      ret.nxt &= ((nxt << emm.len) | (all >> (n - emm.len - 1)));
    if (len + emm.len >= n) {
      bitset<30001> res = nxt & emm.pre;
      if (len < n - 1) res &= (all >> (n - 1 - len));
      if (emm.len < n - 1) res &= (all << (n - 1 - emm.len));
      ret.ans += res.count();
    }
    return ret;
  }
} f[100][100];
inline long long query(long long x) {
  ooxx ret;
  ret.ans = ret.len = 0, ret.nxt.reset(), ret.pre.reset();
  long long now = 0;
  for (int i = tot; i >= 0; --i) {
    for (int j = 1; j <= d; ++j) {
      if (len[i] > x) {
        now += gen[j];
        now %= m;
        break;
      } else {
        ret = (ret + f[i][(gen[j] + now) % m]);
        x -= len[i];
      }
    }
  }
  return ret.ans;
}
int main() {
  d = ri();
  m = ri();
  for (int i = 1; i <= d; ++i) gen[i] = ri();
  n = ri();
  for (int i = 1; i <= n; ++i) b[i] = ri();
  for (int i = 1; i < n; ++i) all.set(i);
  len[tot = 0] = 1;
  for (int i = 0; i < m; ++i) {
    f[0][i].len = 1;
    if (n == 1)
      f[0][i].ans = i <= b[1];
    else {
      for (int j = 1; j <= n - 1; ++j) {
        f[0][i].pre[j] = i <= b[j + 1];
        f[0][i].nxt[j] = i <= b[j];
      }
    }
  }
  long long L = ri(), R = ri();
  for (; len[tot] <= R / d;) {
    ++tot;
    len[tot] = len[tot - 1] * d;
    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]);
  }
  printf("%lld\n", query(R) - query(L + n - 2));
  return 0;
}