#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
long long a[N], b[N];
unordered_map<long long, int> mp;
int main(void) {
  int t;
  t = 1;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    long long x, y, k;
    cin >> x >> k >> y;
    for (int i = 1; i <= n; i++) cin >> a[i], mp[a[i]] = i;
    for (int i = 1; i <= m; i++) cin >> b[i];
    if (m > n)
      cout << -1;
    else {
      long long last = 0, flag = 1;
      for (int i = 1; i <= m; i++) {
        if (mp[b[i]] < last) {
          flag = 0;
          break;
        }
        last = mp[b[i]];
      }
      if (!flag)
        cout << -1;
      else {
        long long ans = 0, pre = 0;
        for (int i = 1; i <= m; i++) {
          long long pos = mp[b[i]];
          long long num = pos - pre - 1;
          long long maxi = max(b[i - 1], b[i]);
          if (!num) {
            pre = pos;
            continue;
          }
          long long num1 = 0, num2 = 0;
          for (int j = pre + 1; j < pos; j++) {
            if (a[j] > maxi)
              num1++;
            else
              num2++;
          }
          if (pos - pre - 1 < k) {
            if (num1) {
              flag = 0;
              break;
            } else
              ans += (pos - pre - 1) * y;
          } else {
            long long cnt1 = y * k, cnt2 = x, temp = pos - pre - 1;
            if (cnt1 <= x) {
              if (num1) {
                ans += x;
                temp -= k;
                ans += temp * y;
              } else {
                ans += temp * y;
              }
            } else {
              ans += (temp / k * x) + (temp % k) * y;
            }
          }
          pre = pos;
        }
        if (!flag)
          cout << -1;
        else {
          if (mp[b[m]] != n) {
            long long pos = mp[b[m]], temp = n - pos;
            long long num1 = 0;
            for (int i = pos + 1; i <= n; i++) {
              if (a[i] > b[m]) num1++;
            }
            if (temp < k) {
              if (num1) {
                flag = 0;
              } else
                ans += temp * y;
            } else {
              long long cnt1 = y * k, cnt2 = x;
              if (cnt1 <= x) {
                if (num1) {
                  ans += x;
                  temp -= k;
                  ans += temp * y;
                } else {
                  ans += temp * y;
                }
              } else {
                ans += (temp / k * x) + (temp % k) * y;
              }
            }
          }
          if (!flag)
            cout << -1;
          else
            cout << ans;
        }
      }
    }
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 int N = 200010;
long long a[N], b[N];
unordered_map<long long, int> mp;
int main(void) {
  int t;
  t = 1;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    long long x, y, k;
    cin >> x >> k >> y;
    for (int i = 1; i <= n; i++) cin >> a[i], mp[a[i]] = i;
    for (int i = 1; i <= m; i++) cin >> b[i];
    if (m > n)
      cout << -1;
    else {
      long long last = 0, flag = 1;
      for (int i = 1; i <= m; i++) {
        if (mp[b[i]] < last) {
          flag = 0;
          break;
        }
        last = mp[b[i]];
      }
      if (!flag)
        cout << -1;
      else {
        long long ans = 0, pre = 0;
        for (int i = 1; i <= m; i++) {
          long long pos = mp[b[i]];
          long long num = pos - pre - 1;
          long long maxi = max(b[i - 1], b[i]);
          if (!num) {
            pre = pos;
            continue;
          }
          long long num1 = 0, num2 = 0;
          for (int j = pre + 1; j < pos; j++) {
            if (a[j] > maxi)
              num1++;
            else
              num2++;
          }
          if (pos - pre - 1 < k) {
            if (num1) {
              flag = 0;
              break;
            } else
              ans += (pos - pre - 1) * y;
          } else {
            long long cnt1 = y * k, cnt2 = x, temp = pos - pre - 1;
            if (cnt1 <= x) {
              if (num1) {
                ans += x;
                temp -= k;
                ans += temp * y;
              } else {
                ans += temp * y;
              }
            } else {
              ans += (temp / k * x) + (temp % k) * y;
            }
          }
          pre = pos;
        }
        if (!flag)
          cout << -1;
        else {
          if (mp[b[m]] != n) {
            long long pos = mp[b[m]], temp = n - pos;
            long long num1 = 0;
            for (int i = pos + 1; i <= n; i++) {
              if (a[i] > b[m]) num1++;
            }
            if (temp < k) {
              if (num1) {
                flag = 0;
              } else
                ans += temp * y;
            } else {
              long long cnt1 = y * k, cnt2 = x;
              if (cnt1 <= x) {
                if (num1) {
                  ans += x;
                  temp -= k;
                  ans += temp * y;
                } else {
                  ans += temp * y;
                }
              } else {
                ans += (temp / k * x) + (temp % k) * y;
              }
            }
          }
          if (!flag)
            cout << -1;
          else
            cout << ans;
        }
      }
    }
  }
  return 0;
}

```