#include <bits/stdc++.h>
using namespace std;
long long n, m, fc, fr, bc;
long long a[200002];
long long b[200002];
vector<vector<long long> > gaps;
int set_gaps() {
  int prev = 1;
  for (int i = 1; i < m + 2; i++) {
    vector<long long> gap;
    int j;
    for (j = prev; j <= n; j++) {
      if (i <= m && a[j] == b[i]) break;
      gap.push_back(j);
    }
    if (j == n + 1 && i < m + 1) return 0;
    if (gap.size() > 0) gaps.push_back(gap);
    prev = j + 1;
  }
  return 1;
}
long long solve() {
  if (set_gaps() == 0) return -1;
  long long res = 0;
  for (int i = 0; i < gaps.size(); i++) {
    if (gaps[i].size() == 1) {
      if (a[gaps[i][0]] < a[gaps[i][0] - 1] ||
          a[gaps[i][0]] < a[gaps[i][0] + 1])
        res = fr == 1 && fc < bc ? res + fc : res + bc;
      else if (fr == 1)
        res += fc;
      else {
        return -1;
      }
    } else {
      long long mx = 0;
      for (int j = 0; j < gaps[i].size(); j++) {
        int g = gaps[i][j];
        mx = max(mx, a[g]);
      }
      if (mx < a[gaps[i][0] - 1] || mx < a[gaps[i][gaps[i].size() - 1] + 1]) {
        if (gaps[i].size() / fr > 0) {
          if (bc * fr < fc) {
            res += gaps[i].size() * bc;
          } else {
            res += (gaps[i].size() / fr) * fc;
            res += bc * (gaps[i].size() - fr * (gaps[i].size() / fr));
          }
        } else {
          res += gaps[i].size() * bc;
        }
      } else {
        if (gaps[i].size() / fr <= 0)
          return -1;
        else {
          if (bc * fr < fc) {
            res += fc + (gaps[i].size() - fr) * bc;
          } else {
            res += (gaps[i].size() / fr) * fc;
            res += bc * (gaps[i].size() - fr * (gaps[i].size() / fr));
          }
        }
      }
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  cin >> fc >> fr >> bc;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  a[0] = -1;
  a[n + 1] = -1;
  cout << solve() << endl;
  return 0;
}

### Prompt

Please formulate 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;
long long n, m, fc, fr, bc;
long long a[200002];
long long b[200002];
vector<vector<long long> > gaps;
int set_gaps() {
  int prev = 1;
  for (int i = 1; i < m + 2; i++) {
    vector<long long> gap;
    int j;
    for (j = prev; j <= n; j++) {
      if (i <= m && a[j] == b[i]) break;
      gap.push_back(j);
    }
    if (j == n + 1 && i < m + 1) return 0;
    if (gap.size() > 0) gaps.push_back(gap);
    prev = j + 1;
  }
  return 1;
}
long long solve() {
  if (set_gaps() == 0) return -1;
  long long res = 0;
  for (int i = 0; i < gaps.size(); i++) {
    if (gaps[i].size() == 1) {
      if (a[gaps[i][0]] < a[gaps[i][0] - 1] ||
          a[gaps[i][0]] < a[gaps[i][0] + 1])
        res = fr == 1 && fc < bc ? res + fc : res + bc;
      else if (fr == 1)
        res += fc;
      else {
        return -1;
      }
    } else {
      long long mx = 0;
      for (int j = 0; j < gaps[i].size(); j++) {
        int g = gaps[i][j];
        mx = max(mx, a[g]);
      }
      if (mx < a[gaps[i][0] - 1] || mx < a[gaps[i][gaps[i].size() - 1] + 1]) {
        if (gaps[i].size() / fr > 0) {
          if (bc * fr < fc) {
            res += gaps[i].size() * bc;
          } else {
            res += (gaps[i].size() / fr) * fc;
            res += bc * (gaps[i].size() - fr * (gaps[i].size() / fr));
          }
        } else {
          res += gaps[i].size() * bc;
        }
      } else {
        if (gaps[i].size() / fr <= 0)
          return -1;
        else {
          if (bc * fr < fc) {
            res += fc + (gaps[i].size() - fr) * bc;
          } else {
            res += (gaps[i].size() / fr) * fc;
            res += bc * (gaps[i].size() - fr * (gaps[i].size() / fr));
          }
        }
      }
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  cin >> fc >> fr >> bc;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  a[0] = -1;
  a[n + 1] = -1;
  cout << solve() << endl;
  return 0;
}

```