#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long t, n, m, x, y, k, ans;
long long a[1000010], b[1000010], pos[1000010], tag[1000010];
int main() {
  ios_base::sync_with_stdio(false);
  t = 1;
  while (t--) {
    ans = 0;
    cin >> n >> m;
    cin >> x >> k >> y;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      pos[a[i]] = i;
      tag[i] = 0;
    }
    int now = 0;
    bool flag = 0;
    for (int i = 1; i <= m; ++i) {
      cin >> b[i];
      if (pos[b[i]] <= now) {
        flag = 1;
      } else {
        now = pos[b[i]];
        tag[pos[b[i]]] = 1;
      }
    }
    tag[n + 1] = 1;
    if (flag) {
      cout << "-1\n";
      continue;
    }
    int per = 0;
    for (int i = 1; i <= n + 1; ++i) {
      if (!tag[i]) continue;
      if (i == per + 1) {
        per = i;
        continue;
      }
      int len = i - per - 1, l = per + 1, r = i - 1;
      int be = 0, eed = 0;
      for (int j = l; j <= r; ++j) {
        if (a[j] > a[i] && a[j] > a[per]) {
          if (be == 0) be = j;
          eed = j;
        }
      }
      if (!be) {
        if (y * k <= x) {
          ans += y * len;
        } else {
          ans += x * (len / k) + y * (len % k);
        }
      } else {
        if (y * k <= x) {
          if (len < k) {
            flag = 1;
            break;
          } else {
            ans += x + y * (len - k);
          }
        } else {
          if (len < k) {
            flag = 1;
            break;
          } else {
            ans += y * (len % k) + x * (len / k);
          }
        }
      }
      per = i;
    }
    if (flag)
      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 long long mod = 1e9 + 7;
long long t, n, m, x, y, k, ans;
long long a[1000010], b[1000010], pos[1000010], tag[1000010];
int main() {
  ios_base::sync_with_stdio(false);
  t = 1;
  while (t--) {
    ans = 0;
    cin >> n >> m;
    cin >> x >> k >> y;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      pos[a[i]] = i;
      tag[i] = 0;
    }
    int now = 0;
    bool flag = 0;
    for (int i = 1; i <= m; ++i) {
      cin >> b[i];
      if (pos[b[i]] <= now) {
        flag = 1;
      } else {
        now = pos[b[i]];
        tag[pos[b[i]]] = 1;
      }
    }
    tag[n + 1] = 1;
    if (flag) {
      cout << "-1\n";
      continue;
    }
    int per = 0;
    for (int i = 1; i <= n + 1; ++i) {
      if (!tag[i]) continue;
      if (i == per + 1) {
        per = i;
        continue;
      }
      int len = i - per - 1, l = per + 1, r = i - 1;
      int be = 0, eed = 0;
      for (int j = l; j <= r; ++j) {
        if (a[j] > a[i] && a[j] > a[per]) {
          if (be == 0) be = j;
          eed = j;
        }
      }
      if (!be) {
        if (y * k <= x) {
          ans += y * len;
        } else {
          ans += x * (len / k) + y * (len % k);
        }
      } else {
        if (y * k <= x) {
          if (len < k) {
            flag = 1;
            break;
          } else {
            ans += x + y * (len - k);
          }
        } else {
          if (len < k) {
            flag = 1;
            break;
          } else {
            ans += y * (len % k) + x * (len / k);
          }
        }
      }
      per = i;
    }
    if (flag)
      cout << "-1\n";
    else
      cout << ans << "\n";
  }
  return 0;
}

```