#include <bits/stdc++.h>
using i64 = long long;
const int maxn = 200000;
inline i64 Min(const i64 &x, const i64 &y) { return x < y ? x : y; }
namespace IOManager {
constexpr int FILESZ = 131072;
char buf[FILESZ];
const char *ibuf = buf, *tbuf = buf;
struct IOManager {
  inline char gc() {
    return (ibuf == tbuf) and
                   (tbuf = (ibuf = buf) + fread(buf, 1, FILESZ, stdin),
                    ibuf == tbuf)
               ? EOF
               : *ibuf++;
  }
  template <typename _Tp>
  inline operator _Tp() {
    _Tp s = 0u;
    char c = gc();
    for (; c < 48; c = gc())
      ;
    for (; c > 47; c = gc()) s = (_Tp)(s * 10u + c - 48u);
    return s;
  }
};
}  // namespace IOManager
IOManager::IOManager io;
int a[maxn + 1], b[maxn + 1], p[maxn + 1];
const int n = io, m = io, x = io, k = io, y = io;
const i64 delk = Min(x, 1ll * y * k);
const bool is = (delk != x);
inline i64 calc(const int &lv, const int &rv, const int &len, const int &mx) {
  if ((len < k) and (mx > lv) and (mx > rv)) throw -1;
  if (len < k) return 1ll * len * y;
  return (len / k) * delk + 1ll * (len % k) * y +
         ((is and (mx > lv) and (mx > rv)) ? (x - delk) : 0);
}
int main() {
  for (int i = 1; i <= n; ++i) a[i] = io;
  for (int i = 1; i <= m; ++i) b[i] = io;
  for (int j = 1, i = 1; j <= m; p[j++] = i) {
    for (; (i <= n) and (a[i] != b[j]); ++i)
      ;
    if (i > n) return puts("-1"), 0;
  }
  try {
    i64 cost = calc(0, b[1], p[1] - 1, *std::max_element(a + 1, a + p[1] + 1)) +
               calc(b[m], 0, n - p[m], *std::max_element(a + p[m], a + n + 1));
    for (int i = 1; i < m; ++i)
      cost += calc(b[i], b[i + 1], p[i + 1] - p[i] - 1,
                   *std::max_element(a + p[i], a + p[i + 1] + 1));
    printf("%lld\n", cost);
  } catch (const int &v) {
    return puts("-1"), 0;
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

There are n warriors in a row. The power of the i-th warrior is a_i. All powers are pairwise distinct.

You have two types of spells which you may cast: 

  1. Fireball: you spend x mana and destroy exactly k consecutive warriors; 
  2. Berserk: you spend y mana, choose two consecutive warriors, and the warrior with greater power destroys the warrior with smaller power. 



For example, let the powers of warriors be [2, 3, 7, 8, 11, 5, 4], and k = 3. If you cast Berserk on warriors with powers 8 and 11, the resulting sequence of powers becomes [2, 3, 7, 11, 5, 4]. Then, for example, if you cast Fireball on consecutive warriors with powers [7, 11, 5], the resulting sequence of powers becomes [2, 3, 4].

You want to turn the current sequence of warriors powers a_1, a_2, ..., a_n into b_1, b_2, ..., b_m. Calculate the minimum amount of mana you need to spend on it.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the length of sequence a and the length of sequence b respectively.

The second line contains three integers x, k, y (1 ≤ x, y, ≤ 10^9; 1 ≤ k ≤ n) — the cost of fireball, the range of fireball and the cost of berserk respectively.

The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n). It is guaranteed that all integers a_i are pairwise distinct.

The fourth line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ n). It is guaranteed that all integers b_i are pairwise distinct.

Output

Print the minimum amount of mana for turning the sequnce a_1, a_2, ..., a_n into b_1, b_2, ..., b_m, or -1 if it is impossible.

Examples

Input


5 2
5 2 3
3 1 4 5 2
3 5


Output


8


Input


4 4
5 1 4
4 3 1 2
2 4 3 1


Output


-1


Input


4 4
2 1 11
1 3 2 4
1 3 2 4


Output


0

### Response

```cpp
#include <bits/stdc++.h>
using i64 = long long;
const int maxn = 200000;
inline i64 Min(const i64 &x, const i64 &y) { return x < y ? x : y; }
namespace IOManager {
constexpr int FILESZ = 131072;
char buf[FILESZ];
const char *ibuf = buf, *tbuf = buf;
struct IOManager {
  inline char gc() {
    return (ibuf == tbuf) and
                   (tbuf = (ibuf = buf) + fread(buf, 1, FILESZ, stdin),
                    ibuf == tbuf)
               ? EOF
               : *ibuf++;
  }
  template <typename _Tp>
  inline operator _Tp() {
    _Tp s = 0u;
    char c = gc();
    for (; c < 48; c = gc())
      ;
    for (; c > 47; c = gc()) s = (_Tp)(s * 10u + c - 48u);
    return s;
  }
};
}  // namespace IOManager
IOManager::IOManager io;
int a[maxn + 1], b[maxn + 1], p[maxn + 1];
const int n = io, m = io, x = io, k = io, y = io;
const i64 delk = Min(x, 1ll * y * k);
const bool is = (delk != x);
inline i64 calc(const int &lv, const int &rv, const int &len, const int &mx) {
  if ((len < k) and (mx > lv) and (mx > rv)) throw -1;
  if (len < k) return 1ll * len * y;
  return (len / k) * delk + 1ll * (len % k) * y +
         ((is and (mx > lv) and (mx > rv)) ? (x - delk) : 0);
}
int main() {
  for (int i = 1; i <= n; ++i) a[i] = io;
  for (int i = 1; i <= m; ++i) b[i] = io;
  for (int j = 1, i = 1; j <= m; p[j++] = i) {
    for (; (i <= n) and (a[i] != b[j]); ++i)
      ;
    if (i > n) return puts("-1"), 0;
  }
  try {
    i64 cost = calc(0, b[1], p[1] - 1, *std::max_element(a + 1, a + p[1] + 1)) +
               calc(b[m], 0, n - p[m], *std::max_element(a + p[m], a + n + 1));
    for (int i = 1; i < m; ++i)
      cost += calc(b[i], b[i + 1], p[i + 1] - p[i] - 1,
                   *std::max_element(a + p[i], a + p[i + 1] + 1));
    printf("%lld\n", cost);
  } catch (const int &v) {
    return puts("-1"), 0;
  }
  return 0;
}

```