#include <bits/stdc++.h>
using namespace std;
const int PRIME = 151;
const long long MOD = 1e9 + 7;
const int MAXN = 2e5 + 5;
const int INF = 0x3f3f3f3f;
long long t, n, m, x, y, k, a[MAXN], b[MAXN];
int main() {
  {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  cin >> n >> m >> x >> k >> y;
  int pos[MAXN] = {0};
  for (int i = 1; i <= n; ++i) cin >> a[i], pos[a[i]] = i;
  int c = 0;
  for (int i = 1; i <= m; ++i) {
    cin >> b[i];
    if (b[i] == a[i]) c++;
  }
  if (c == n) {
    cout << 0 << '\n';
    return 0;
  }
  long long ans = 0;
  for (int i = 1; i <= m + 1; ++i) {
    int l = pos[b[i - 1]], r = (i == m + 1 ? n + 1 : pos[b[i]]);
    if (r < l) {
      cout << -1 << '\n';
      return 0;
    }
    if (r - l <= 1) continue;
    long long Max = *max_element(a + l + 1, a + r), cnt = r - l - 1;
    if (b[i] > Max || b[i - 1] > Max) {
      if (cnt >= k) {
        ans += min(cnt / k * x + (cnt % k) * y, cnt * y);
      } else {
        ans += cnt * y;
      }
    } else {
      if (cnt >= k) {
        long long tt = cnt - k;
        ans += min(tt / k * x + (tt % k) * y, tt * y) + x;
      } else {
        cout << -1 << '\n';
        return 0;
      }
    }
  }
  if (!ans)
    cout << -1 << '\n';
  else
    cout << ans << '\n';
  return 0;
}

### Prompt

Generate 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 PRIME = 151;
const long long MOD = 1e9 + 7;
const int MAXN = 2e5 + 5;
const int INF = 0x3f3f3f3f;
long long t, n, m, x, y, k, a[MAXN], b[MAXN];
int main() {
  {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  cin >> n >> m >> x >> k >> y;
  int pos[MAXN] = {0};
  for (int i = 1; i <= n; ++i) cin >> a[i], pos[a[i]] = i;
  int c = 0;
  for (int i = 1; i <= m; ++i) {
    cin >> b[i];
    if (b[i] == a[i]) c++;
  }
  if (c == n) {
    cout << 0 << '\n';
    return 0;
  }
  long long ans = 0;
  for (int i = 1; i <= m + 1; ++i) {
    int l = pos[b[i - 1]], r = (i == m + 1 ? n + 1 : pos[b[i]]);
    if (r < l) {
      cout << -1 << '\n';
      return 0;
    }
    if (r - l <= 1) continue;
    long long Max = *max_element(a + l + 1, a + r), cnt = r - l - 1;
    if (b[i] > Max || b[i - 1] > Max) {
      if (cnt >= k) {
        ans += min(cnt / k * x + (cnt % k) * y, cnt * y);
      } else {
        ans += cnt * y;
      }
    } else {
      if (cnt >= k) {
        long long tt = cnt - k;
        ans += min(tt / k * x + (tt % k) * y, tt * y) + x;
      } else {
        cout << -1 << '\n';
        return 0;
      }
    }
  }
  if (!ans)
    cout << -1 << '\n';
  else
    cout << ans << '\n';
  return 0;
}

```