#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long int i, j, n, s = 0, c = 0, x, k, y, m, z, ans = 0;
  cin >> n >> m;
  cin >> x >> z >> y;
  vector<long long int> a(n), b(m);
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < m; i++) {
    cin >> b[i];
  }
  if (n == m) {
    for (i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        cout << -1;
        return;
      }
    }
    cout << 0;
    return;
  }
  if (m > n) {
    cout << -1;
    return;
  }
  i = 0;
  j = 0;
  k = 0;
  long long int ma = INT_MIN, l;
  while (i < n && j < m) {
    if (a[i] != b[j]) {
      ma = max(ma, a[i]);
      i++;
    } else {
      l = i - k;
      if (l == 0) {
        i++;
        j++;
        k = i;
        continue;
      }
      long long int maxel;
      if (k > 0) {
        maxel = max(a[k - 1], a[i]);
      } else {
        maxel = a[i];
      }
      if (l < z) {
        if (maxel < ma) {
          cout << -1;
          return;
        }
        ans += l * y;
      } else {
        if (y * z < x && maxel > ma) {
          ans += l * y;
        } else if (y * z < x) {
          l = l - z;
          ans += x + l * y;
        } else {
          long long int rem = l % z;
          long long int gro = l / z;
          ans += gro * x + rem * y;
        }
      }
      j++;
      i++;
      k = i;
      ma = INT_MIN;
    }
  }
  l = n - k;
  for (long long int q = i; q < n; q++) {
    ma = max(ma, a[q]);
  }
  if (l > 0) {
    long long int maxel = a[k - 1];
    if (l < z) {
      if (maxel < ma) {
        cout << -1;
        return;
      }
      ans += l * y;
    } else {
      if (y * z < x && maxel > ma) {
        ans += l * y;
      } else if (y * z < x) {
        l = l - z;
        ans += x + l * y;
      } else {
        long long int rem = l % z;
        long long int gro = l / z;
        ans += gro * x + rem * y;
      }
    }
    j++;
    i++;
    k = i;
    ma = INT_MIN;
  }
  cout << ans;
}
int main() {
  long long int t;
  t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
}

### 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;
void solve() {
  long long int i, j, n, s = 0, c = 0, x, k, y, m, z, ans = 0;
  cin >> n >> m;
  cin >> x >> z >> y;
  vector<long long int> a(n), b(m);
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < m; i++) {
    cin >> b[i];
  }
  if (n == m) {
    for (i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        cout << -1;
        return;
      }
    }
    cout << 0;
    return;
  }
  if (m > n) {
    cout << -1;
    return;
  }
  i = 0;
  j = 0;
  k = 0;
  long long int ma = INT_MIN, l;
  while (i < n && j < m) {
    if (a[i] != b[j]) {
      ma = max(ma, a[i]);
      i++;
    } else {
      l = i - k;
      if (l == 0) {
        i++;
        j++;
        k = i;
        continue;
      }
      long long int maxel;
      if (k > 0) {
        maxel = max(a[k - 1], a[i]);
      } else {
        maxel = a[i];
      }
      if (l < z) {
        if (maxel < ma) {
          cout << -1;
          return;
        }
        ans += l * y;
      } else {
        if (y * z < x && maxel > ma) {
          ans += l * y;
        } else if (y * z < x) {
          l = l - z;
          ans += x + l * y;
        } else {
          long long int rem = l % z;
          long long int gro = l / z;
          ans += gro * x + rem * y;
        }
      }
      j++;
      i++;
      k = i;
      ma = INT_MIN;
    }
  }
  l = n - k;
  for (long long int q = i; q < n; q++) {
    ma = max(ma, a[q]);
  }
  if (l > 0) {
    long long int maxel = a[k - 1];
    if (l < z) {
      if (maxel < ma) {
        cout << -1;
        return;
      }
      ans += l * y;
    } else {
      if (y * z < x && maxel > ma) {
        ans += l * y;
      } else if (y * z < x) {
        l = l - z;
        ans += x + l * y;
      } else {
        long long int rem = l % z;
        long long int gro = l / z;
        ans += gro * x + rem * y;
      }
    }
    j++;
    i++;
    k = i;
    ma = INT_MIN;
  }
  cout << ans;
}
int main() {
  long long int t;
  t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
}

```