#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long T = 1;
  while (T--) {
    long long n, m;
    cin >> n >> m;
    long long x, k, y;
    cin >> x >> k >> y;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<long long> b(n);
    for (int i = 0; i < m; i++) {
      cin >> b[i];
    }
    bool f0 = true;
    long long i1 = 0, i2 = 0;
    long long cmx = -1;
    while (i1 < n && a[i1] != b[i2]) {
      cmx = max(cmx, a[i1]);
      i1++;
    }
    if (i1 == n || (cmx > a[i1] && i1 < k)) {
      cout << -1;
      return 0;
    }
    long long ans = 0;
    if (i1 >= k) {
      if (x < k * y) {
        ans += ((i1 / k) * x) + ((i1 % k) * y);
      } else {
        if (cmx > a[i1]) {
          ans += x + (i1 - k) * y;
        } else {
          ans += i1 * y;
        }
      }
    } else {
      ans += i1 * y;
    }
    long long j1 = i1 + 1, j2 = i2 + 1;
    while (j1 < n && j2 < m) {
      cmx = -1;
      while (j1 < n && a[j1] != b[j2]) {
        cmx = max(cmx, a[j1]);
        j1++;
      }
      if (j1 == n || (cmx > a[j1] && cmx > a[i1] && j1 - i1 - 1 < k)) {
        cout << -1;
        return 0;
      }
      if (j1 - i1 - 1 >= k) {
        if (x < k * y) {
          ans += (((j1 - i1 - 1) / k) * x) + (((j1 - i1 - 1) % k) * y);
        } else {
          if (cmx > a[j1] && cmx > a[i1]) {
            ans += x + (j1 - i1 - 1 - k) * y;
          } else {
            ans += (j1 - i1 - 1) * y;
          }
        }
      } else {
        ans += (j1 - i1 - 1) * y;
      }
      i1 = j1;
      j1++;
      i2 = j2;
      j2++;
    }
    if (j2 < m && j1 == n) {
      cout << -1;
      return 0;
    }
    if (j1 < n) {
      cmx = a[j1];
      j1++;
      while (j1 < n) {
        cmx = max(cmx, a[j1]);
        j1++;
      }
      if (cmx > a[i1] && j1 - i1 - 1 < k) {
        cout << -1;
        return 0;
      }
      if (j1 - i1 - 1 >= k) {
        if (x < k * y) {
          ans += (((j1 - i1 - 1) / k) * x) + (((j1 - i1 - 1) % k) * y);
        } else {
          if (cmx > a[j1] && cmx > a[i1]) {
            ans += x + (j1 - i1 - 1 - k) * y;
          } else {
            ans += (j1 - i1 - 1) * y;
          }
        }
      } else {
        ans += (j1 - i1 - 1) * y;
      }
    }
    cout << ans;
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long T = 1;
  while (T--) {
    long long n, m;
    cin >> n >> m;
    long long x, k, y;
    cin >> x >> k >> y;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<long long> b(n);
    for (int i = 0; i < m; i++) {
      cin >> b[i];
    }
    bool f0 = true;
    long long i1 = 0, i2 = 0;
    long long cmx = -1;
    while (i1 < n && a[i1] != b[i2]) {
      cmx = max(cmx, a[i1]);
      i1++;
    }
    if (i1 == n || (cmx > a[i1] && i1 < k)) {
      cout << -1;
      return 0;
    }
    long long ans = 0;
    if (i1 >= k) {
      if (x < k * y) {
        ans += ((i1 / k) * x) + ((i1 % k) * y);
      } else {
        if (cmx > a[i1]) {
          ans += x + (i1 - k) * y;
        } else {
          ans += i1 * y;
        }
      }
    } else {
      ans += i1 * y;
    }
    long long j1 = i1 + 1, j2 = i2 + 1;
    while (j1 < n && j2 < m) {
      cmx = -1;
      while (j1 < n && a[j1] != b[j2]) {
        cmx = max(cmx, a[j1]);
        j1++;
      }
      if (j1 == n || (cmx > a[j1] && cmx > a[i1] && j1 - i1 - 1 < k)) {
        cout << -1;
        return 0;
      }
      if (j1 - i1 - 1 >= k) {
        if (x < k * y) {
          ans += (((j1 - i1 - 1) / k) * x) + (((j1 - i1 - 1) % k) * y);
        } else {
          if (cmx > a[j1] && cmx > a[i1]) {
            ans += x + (j1 - i1 - 1 - k) * y;
          } else {
            ans += (j1 - i1 - 1) * y;
          }
        }
      } else {
        ans += (j1 - i1 - 1) * y;
      }
      i1 = j1;
      j1++;
      i2 = j2;
      j2++;
    }
    if (j2 < m && j1 == n) {
      cout << -1;
      return 0;
    }
    if (j1 < n) {
      cmx = a[j1];
      j1++;
      while (j1 < n) {
        cmx = max(cmx, a[j1]);
        j1++;
      }
      if (cmx > a[i1] && j1 - i1 - 1 < k) {
        cout << -1;
        return 0;
      }
      if (j1 - i1 - 1 >= k) {
        if (x < k * y) {
          ans += (((j1 - i1 - 1) / k) * x) + (((j1 - i1 - 1) % k) * y);
        } else {
          if (cmx > a[j1] && cmx > a[i1]) {
            ans += x + (j1 - i1 - 1 - k) * y;
          } else {
            ans += (j1 - i1 - 1) * y;
          }
        }
      } else {
        ans += (j1 - i1 - 1) * y;
      }
    }
    cout << ans;
  }
}

```