#include <bits/stdc++.h>
using namespace std;
inline int read() {
  char ch = getchar();
  int f = 1, x = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n;
int a[200005], b[200005];
int m, x, y, k, pos[200005];
long long calc(int l, int r) {
  if (r - l <= 1) return 0;
  int mx = 0;
  for (int i = l + 1; i < r; i++) mx = max(mx, a[i]);
  if (mx > max(a[l], a[r]) && r - l - 1 < k) return -1;
  if (r - l - 1 < k) return (long long)y * (r - l - 1);
  if (x > (long long)y * k) {
    if (mx > max(a[l], a[r])) return (long long)y * (r - l - 1 - k) + x;
    return (long long)y * (r - l - 1);
  } else {
    return (long long)x * (long long)((r - l - 1) / k) +
           (long long)y * ((r - l - 1) % k);
  }
}
void solve() {
  n = read();
  m = read();
  x = read();
  k = read();
  y = read();
  long long ans = 0;
  for (int i = 1; i <= n; i++) a[i] = read(), pos[a[i]] = i;
  for (int i = 1; i <= m; i++) b[i] = read();
  for (int i = 1; i < m; i++)
    if (pos[b[i]] > pos[b[i + 1]]) {
      printf("-1");
      return;
    }
  b[m + 1] = n + 1;
  pos[n + 1] = n + 1;
  for (int i = 0; i <= m; i++) {
    long long co = calc(pos[b[i]], pos[b[i + 1]]);
    if (co == -1) {
      printf("-1");
      return;
    }
    ans += co;
  }
  cout << ans;
}
int main() {
  solve();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 namespace std;
inline int read() {
  char ch = getchar();
  int f = 1, x = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n;
int a[200005], b[200005];
int m, x, y, k, pos[200005];
long long calc(int l, int r) {
  if (r - l <= 1) return 0;
  int mx = 0;
  for (int i = l + 1; i < r; i++) mx = max(mx, a[i]);
  if (mx > max(a[l], a[r]) && r - l - 1 < k) return -1;
  if (r - l - 1 < k) return (long long)y * (r - l - 1);
  if (x > (long long)y * k) {
    if (mx > max(a[l], a[r])) return (long long)y * (r - l - 1 - k) + x;
    return (long long)y * (r - l - 1);
  } else {
    return (long long)x * (long long)((r - l - 1) / k) +
           (long long)y * ((r - l - 1) % k);
  }
}
void solve() {
  n = read();
  m = read();
  x = read();
  k = read();
  y = read();
  long long ans = 0;
  for (int i = 1; i <= n; i++) a[i] = read(), pos[a[i]] = i;
  for (int i = 1; i <= m; i++) b[i] = read();
  for (int i = 1; i < m; i++)
    if (pos[b[i]] > pos[b[i + 1]]) {
      printf("-1");
      return;
    }
  b[m + 1] = n + 1;
  pos[n + 1] = n + 1;
  for (int i = 0; i <= m; i++) {
    long long co = calc(pos[b[i]], pos[b[i + 1]]);
    if (co == -1) {
      printf("-1");
      return;
    }
    ans += co;
  }
  cout << ans;
}
int main() {
  solve();
  return 0;
}

```