#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
int a[N], b[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  int x, k, y;
  cin >> x >> k >> y;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  a[n + 1] = b[m + 1] = -1;
  long long ans = 0;
  for (int i = 1, j = 0; i <= m + 1; i++) {
    int l = j + 1, r = j + 1, len = 0, maxx = -1;
    while (r <= n + 1 && a[r] != b[i]) maxx = max(maxx, a[r]), r++;
    if (r > n + 1) {
      ans = -1;
      break;
    }
    len = r - l;
    if (len < k) {
      if (maxx > max(a[l - 1], a[r])) {
        ans = -1;
        break;
      } else
        ans += 1ll * y * len;
    } else {
      if (maxx > max(a[l - 1], a[r])) {
        ans += 1ll * x;
        len -= k;
        if (x < 1ll * k * y)
          ans += 1ll * (len / k) * x + 1ll * (len % k) * y;
        else
          ans += 1ll * y * len;
      } else {
        if (x < 1ll * k * y)
          ans += 1ll * (len / k) * x + 1ll * (len % k) * y;
        else
          ans += 1ll * y * len;
      }
    }
    if (a[r] != b[i])
      j = r + 1;
    else
      j = r;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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 namespace std;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
int a[N], b[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  int x, k, y;
  cin >> x >> k >> y;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  a[n + 1] = b[m + 1] = -1;
  long long ans = 0;
  for (int i = 1, j = 0; i <= m + 1; i++) {
    int l = j + 1, r = j + 1, len = 0, maxx = -1;
    while (r <= n + 1 && a[r] != b[i]) maxx = max(maxx, a[r]), r++;
    if (r > n + 1) {
      ans = -1;
      break;
    }
    len = r - l;
    if (len < k) {
      if (maxx > max(a[l - 1], a[r])) {
        ans = -1;
        break;
      } else
        ans += 1ll * y * len;
    } else {
      if (maxx > max(a[l - 1], a[r])) {
        ans += 1ll * x;
        len -= k;
        if (x < 1ll * k * y)
          ans += 1ll * (len / k) * x + 1ll * (len % k) * y;
        else
          ans += 1ll * y * len;
      } else {
        if (x < 1ll * k * y)
          ans += 1ll * (len / k) * x + 1ll * (len % k) * y;
        else
          ans += 1ll * y * len;
      }
    }
    if (a[r] != b[i])
      j = r + 1;
    else
      j = r;
  }
  cout << ans << endl;
  return 0;
}

```