#include <bits/stdc++.h>
using namespace std;
using namespace std;
struct sch {
  long long int s, e, t;
};
bool sortinrev(const struct sch &a, const struct sch &b) { return (a.s < b.s); }
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m;
  cin >> n >> m;
  long long int x, k, y;
  cin >> x >> k >> y;
  long long int a[n];
  long long int b[m];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long int i = 0; i < m; i++) {
    cin >> b[i];
  }
  int fl = 1;
  vector<vector<long long int> > rem;
  vector<long long int> tmp;
  long long int j = 0;
  for (long long int i = 0; i < n; i++) {
    if (j < m && a[i] == b[j]) {
      if (tmp.size()) {
        rem.push_back(tmp);
        tmp.clear();
      }
      j++;
    } else {
      tmp.push_back(i);
    }
  }
  if (tmp.size()) {
    rem.push_back(tmp);
  }
  if (j < m) {
    cout << -1 << endl;
    return 0;
  }
  long long int ans = 0;
  for (long long int i = 0; i < rem.size(); i++) {
    tmp = rem[i];
    if (tmp.size() < k) {
      long long int sz = tmp.size();
      long long int pos1 = tmp[0], pos2 = tmp[sz - 1];
      long long int ma = a[pos1];
      for (j = pos1; j <= pos2; j++) {
        ma = max(ma, a[j]);
      }
      if ((pos1 && a[pos1 - 1] > ma) || (pos2 < n - 1 && a[pos2 + 1] > ma)) {
        ans += y * sz;
      } else {
        fl = 0;
        break;
      }
    } else {
      if (x <= k * y) {
        long long int sz = tmp.size();
        ans += (sz / k) * x;
        ans += (sz % k) * y;
      } else {
        long long int sz = tmp.size();
        long long int pos1 = tmp[0], pos2 = tmp[sz - 1];
        long long int ma = a[pos1];
        for (j = pos1; j <= pos2; j++) {
          ma = max(ma, a[j]);
        }
        if ((pos1 && a[pos1 - 1] > ma) || (pos2 < n - 1 && a[pos2 + 1] > ma)) {
          ans += y * sz;
        } else {
          ans += x;
          ans += y * (sz - k);
        }
      }
    }
  }
  if (fl == 0) {
    cout << -1 << endl;
  } else {
    cout << ans << endl;
  }
}

### 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;
using namespace std;
struct sch {
  long long int s, e, t;
};
bool sortinrev(const struct sch &a, const struct sch &b) { return (a.s < b.s); }
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m;
  cin >> n >> m;
  long long int x, k, y;
  cin >> x >> k >> y;
  long long int a[n];
  long long int b[m];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long int i = 0; i < m; i++) {
    cin >> b[i];
  }
  int fl = 1;
  vector<vector<long long int> > rem;
  vector<long long int> tmp;
  long long int j = 0;
  for (long long int i = 0; i < n; i++) {
    if (j < m && a[i] == b[j]) {
      if (tmp.size()) {
        rem.push_back(tmp);
        tmp.clear();
      }
      j++;
    } else {
      tmp.push_back(i);
    }
  }
  if (tmp.size()) {
    rem.push_back(tmp);
  }
  if (j < m) {
    cout << -1 << endl;
    return 0;
  }
  long long int ans = 0;
  for (long long int i = 0; i < rem.size(); i++) {
    tmp = rem[i];
    if (tmp.size() < k) {
      long long int sz = tmp.size();
      long long int pos1 = tmp[0], pos2 = tmp[sz - 1];
      long long int ma = a[pos1];
      for (j = pos1; j <= pos2; j++) {
        ma = max(ma, a[j]);
      }
      if ((pos1 && a[pos1 - 1] > ma) || (pos2 < n - 1 && a[pos2 + 1] > ma)) {
        ans += y * sz;
      } else {
        fl = 0;
        break;
      }
    } else {
      if (x <= k * y) {
        long long int sz = tmp.size();
        ans += (sz / k) * x;
        ans += (sz % k) * y;
      } else {
        long long int sz = tmp.size();
        long long int pos1 = tmp[0], pos2 = tmp[sz - 1];
        long long int ma = a[pos1];
        for (j = pos1; j <= pos2; j++) {
          ma = max(ma, a[j]);
        }
        if ((pos1 && a[pos1 - 1] > ma) || (pos2 < n - 1 && a[pos2 + 1] > ma)) {
          ans += y * sz;
        } else {
          ans += x;
          ans += y * (sz - k);
        }
      }
    }
  }
  if (fl == 0) {
    cout << -1 << endl;
  } else {
    cout << ans << endl;
  }
}

```