#include <bits/stdc++.h>
using namespace std;
long long x, k, y, cnt;
const int mod = 2e3;
const int Max = 2e5 + 5;
long long data[Max], tdata[Max];
bool vis[Max];
int main() {
  int n, m;
  cin >> n >> m;
  cin >> x >> k >> y;
  for (int i = 1; i <= n; i++) cin >> data[i], vis[i] = 0;
  for (int i = 1; i <= m; i++) cin >> tdata[i];
  cnt = 1;
  for (int i = 1; i <= n; i++) {
    if (data[i] == tdata[cnt]) {
      cnt++;
      vis[i] = 1;
      continue;
    }
  }
  if (cnt < m)
    cout << -1 << '\n';
  else {
    long long ans = 0, maxn = 0, lastone = 0;
    bool flag = 0;
    vis[n + 1] = 1;
    cnt = 0;
    for (int i = 1; i <= n + 1; i++) {
      if (vis[i]) {
        if (!cnt) {
          lastone = data[i];
          continue;
        }
        long long L = (maxn < lastone || maxn < data[i]) ? 0 : 1, R = cnt / k;
        if (L > R) {
          flag = 1;
          break;
        }
        long long team;
        if (k * y - x > 0)
          team = max(L, min(R, (cnt * y) / (k * y - x)));
        else
          team = L;
        ans += team * x;
        ans += (cnt - team * k) * y;
        cnt = 0;
        maxn = 0;
        lastone = data[i];
      } else {
        cnt++;
        maxn = max(maxn, data[i]);
      }
    }
    if (!flag)
      cout << ans << '\n';
    else
      cout << -1 << '\n';
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
long long x, k, y, cnt;
const int mod = 2e3;
const int Max = 2e5 + 5;
long long data[Max], tdata[Max];
bool vis[Max];
int main() {
  int n, m;
  cin >> n >> m;
  cin >> x >> k >> y;
  for (int i = 1; i <= n; i++) cin >> data[i], vis[i] = 0;
  for (int i = 1; i <= m; i++) cin >> tdata[i];
  cnt = 1;
  for (int i = 1; i <= n; i++) {
    if (data[i] == tdata[cnt]) {
      cnt++;
      vis[i] = 1;
      continue;
    }
  }
  if (cnt < m)
    cout << -1 << '\n';
  else {
    long long ans = 0, maxn = 0, lastone = 0;
    bool flag = 0;
    vis[n + 1] = 1;
    cnt = 0;
    for (int i = 1; i <= n + 1; i++) {
      if (vis[i]) {
        if (!cnt) {
          lastone = data[i];
          continue;
        }
        long long L = (maxn < lastone || maxn < data[i]) ? 0 : 1, R = cnt / k;
        if (L > R) {
          flag = 1;
          break;
        }
        long long team;
        if (k * y - x > 0)
          team = max(L, min(R, (cnt * y) / (k * y - x)));
        else
          team = L;
        ans += team * x;
        ans += (cnt - team * k) * y;
        cnt = 0;
        maxn = 0;
        lastone = data[i];
      } else {
        cnt++;
        maxn = max(maxn, data[i]);
      }
    }
    if (!flag)
      cout << ans << '\n';
    else
      cout << -1 << '\n';
  }
  return 0;
}

```