#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, mm;
  cin >> n >> mm;
  long long int x, k, y;
  cin >> x >> k >> y;
  long long int a[n];
  long long int b[n];
  map<long long int, long long int> m;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]] = i;
  }
  long long int f = 0;
  for (long long int i = 0; i < mm; i++) {
    cin >> b[i];
    if (m.find(b[i]) == m.end()) {
      f = 1;
    } else if (i > m[b[i]]) {
      f = 1;
    }
  }
  if (f == 1)
    cout << "-1"
         << "\n";
  else if (n == mm)
    cout << "0"
         << "\n";
  else {
    long long int ct = 0;
    long long int ff = 0;
    long long int h = 0;
    for (long long int i = 0; i < mm; i++) {
      long long int f = 0;
      if (i == 0) {
        long long int sum = 0;
        long long int sum2 = 0;
        long long int kk = 0;
        long long int maxx = b[0];
        for (long long int j = 0; j < m[b[0]]; j++) {
          if (a[j] >= maxx) {
            sum++;
          } else
            sum2++;
        }
        if (x > k * y) {
          if (sum > 0 && sum + sum2 < k) {
            h = 1;
            break;
          } else if (sum > 0) {
            ct += x;
            ct += (sum + sum2 - k) * y;
          } else {
            ct += (sum + sum2) * y;
          }
        } else {
          long long int s1 = (sum + sum2) / k;
          long long int s2 = (sum + sum2) - s1 * k;
          ct += s1 * x;
          ct += s2 * y;
        }
      } else {
        long long int sum = 0;
        long long int sum2 = 0;
        long long int kk = 0;
        long long int maxx = max(b[i - 1], b[i]);
        for (long long int j = m[b[i - 1]] + 1; j < m[b[i]]; j++) {
          if (a[j] >= maxx)
            sum++;
          else
            sum2++;
        }
        if (sum > 0 && sum + sum2 < k) {
          h = 1;
          break;
        } else if (x > k * y) {
          if (sum > 0) {
            ct += x;
            ct += (sum + sum2 - k) * y;
          } else {
            ct += (sum + sum2) * y;
          }
        } else {
          long long int s1 = (sum + sum2) / k;
          long long int s2 = (sum + sum2) - s1 * k;
          ct += s1 * x;
          ct += s2 * y;
        }
      }
    }
    long long int sum = 0;
    long long int sum2 = 0;
    long long int kk = 0;
    long long int maxx = b[mm - 1];
    for (long long int j = m[b[mm - 1]] + 1; j < n; j++) {
      if (a[j] >= maxx)
        sum++;
      else
        sum2++;
    }
    if (x > k * y) {
      if (sum > 0 && sum + sum2 < k) {
        h = 1;
      } else if (sum > 0) {
        ct += x;
        ct += (sum + sum2 - k) * y;
      } else {
        ct += (sum + sum2) * y;
      }
    } else {
      long long int s1 = (sum + sum2) / k;
      long long int s2 = (sum + sum2) - s1 * k;
      ct += s1 * x;
      ct += s2 * y;
    }
    if (h == 1)
      cout << "-1"
           << "\n";
    else
      cout << ct << "\n";
  }
  return 0;
}

### Prompt

Your task is to create 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, mm;
  cin >> n >> mm;
  long long int x, k, y;
  cin >> x >> k >> y;
  long long int a[n];
  long long int b[n];
  map<long long int, long long int> m;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]] = i;
  }
  long long int f = 0;
  for (long long int i = 0; i < mm; i++) {
    cin >> b[i];
    if (m.find(b[i]) == m.end()) {
      f = 1;
    } else if (i > m[b[i]]) {
      f = 1;
    }
  }
  if (f == 1)
    cout << "-1"
         << "\n";
  else if (n == mm)
    cout << "0"
         << "\n";
  else {
    long long int ct = 0;
    long long int ff = 0;
    long long int h = 0;
    for (long long int i = 0; i < mm; i++) {
      long long int f = 0;
      if (i == 0) {
        long long int sum = 0;
        long long int sum2 = 0;
        long long int kk = 0;
        long long int maxx = b[0];
        for (long long int j = 0; j < m[b[0]]; j++) {
          if (a[j] >= maxx) {
            sum++;
          } else
            sum2++;
        }
        if (x > k * y) {
          if (sum > 0 && sum + sum2 < k) {
            h = 1;
            break;
          } else if (sum > 0) {
            ct += x;
            ct += (sum + sum2 - k) * y;
          } else {
            ct += (sum + sum2) * y;
          }
        } else {
          long long int s1 = (sum + sum2) / k;
          long long int s2 = (sum + sum2) - s1 * k;
          ct += s1 * x;
          ct += s2 * y;
        }
      } else {
        long long int sum = 0;
        long long int sum2 = 0;
        long long int kk = 0;
        long long int maxx = max(b[i - 1], b[i]);
        for (long long int j = m[b[i - 1]] + 1; j < m[b[i]]; j++) {
          if (a[j] >= maxx)
            sum++;
          else
            sum2++;
        }
        if (sum > 0 && sum + sum2 < k) {
          h = 1;
          break;
        } else if (x > k * y) {
          if (sum > 0) {
            ct += x;
            ct += (sum + sum2 - k) * y;
          } else {
            ct += (sum + sum2) * y;
          }
        } else {
          long long int s1 = (sum + sum2) / k;
          long long int s2 = (sum + sum2) - s1 * k;
          ct += s1 * x;
          ct += s2 * y;
        }
      }
    }
    long long int sum = 0;
    long long int sum2 = 0;
    long long int kk = 0;
    long long int maxx = b[mm - 1];
    for (long long int j = m[b[mm - 1]] + 1; j < n; j++) {
      if (a[j] >= maxx)
        sum++;
      else
        sum2++;
    }
    if (x > k * y) {
      if (sum > 0 && sum + sum2 < k) {
        h = 1;
      } else if (sum > 0) {
        ct += x;
        ct += (sum + sum2 - k) * y;
      } else {
        ct += (sum + sum2) * y;
      }
    } else {
      long long int s1 = (sum + sum2) / k;
      long long int s2 = (sum + sum2) - s1 * k;
      ct += s1 * x;
      ct += s2 * y;
    }
    if (h == 1)
      cout << "-1"
           << "\n";
    else
      cout << ct << "\n";
  }
  return 0;
}

```