#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long double pi = 3.14159265359;
const long long N = 100005;
long long a[N][35];
long long n;
long long b[35][2];
long long ch = 0;
long long power(long long a, long long b) {
  long long res = 1;
  a %= mod;
  while (b > 0) {
    if (b % 2 == 1) {
      res = (res * a) % mod;
    }
    b /= 2;
    a = (a * a) % mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  long long s = x;
  long long r = y;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long b[m];
  for (long long i = 0; i < m; i++) cin >> b[i];
  long long u = 0;
  if (r * k < s) u = 1;
  long long j = 0;
  vector<pair<long long, long long>> p;
  for (long long i = 0; i < n; i++) {
    if ((j < m && a[i] != b[j]) || j == m) {
      p.push_back(make_pair(a[i], i));
    } else {
      j++;
    }
  }
  if (j != m) {
    cout << "-1";
    return 0;
  }
  long long ans = 0;
  for (long long i = 0; i < p.size();) {
    long long x = p[i].second;
    long long y = 0;
    long long d = 0;
    long long c = 0;
    while (i < p.size()) {
      if (p[i].second != x + y) {
        break;
      } else {
        d = max(p[i].first, d);
        c++;
      }
      i++;
      y++;
    }
    long long prev = x - 1;
    long long nxt = x + y;
    long long t = 0;
    if (prev >= 0) t = a[prev];
    if (nxt < n) t = max(t, a[nxt]);
    long long g = c / k;
    long long e = c % k;
    if (g == 0) {
      if (t < d) {
        cout << "-1";
        return 0;
      } else {
        ans += c * r;
      }
    } else {
      if (u == 1) {
        long long ans1;
        if (t >= d)
          ans += c * r;
        else
          ans += ((c - k) * r + s);
      } else {
        long long ans1 = (e * r + g * s);
        ans += ans1;
      }
    }
  }
  cout << ans << "\n";
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 long long mod = 1e9 + 7;
const long double pi = 3.14159265359;
const long long N = 100005;
long long a[N][35];
long long n;
long long b[35][2];
long long ch = 0;
long long power(long long a, long long b) {
  long long res = 1;
  a %= mod;
  while (b > 0) {
    if (b % 2 == 1) {
      res = (res * a) % mod;
    }
    b /= 2;
    a = (a * a) % mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  long long s = x;
  long long r = y;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long b[m];
  for (long long i = 0; i < m; i++) cin >> b[i];
  long long u = 0;
  if (r * k < s) u = 1;
  long long j = 0;
  vector<pair<long long, long long>> p;
  for (long long i = 0; i < n; i++) {
    if ((j < m && a[i] != b[j]) || j == m) {
      p.push_back(make_pair(a[i], i));
    } else {
      j++;
    }
  }
  if (j != m) {
    cout << "-1";
    return 0;
  }
  long long ans = 0;
  for (long long i = 0; i < p.size();) {
    long long x = p[i].second;
    long long y = 0;
    long long d = 0;
    long long c = 0;
    while (i < p.size()) {
      if (p[i].second != x + y) {
        break;
      } else {
        d = max(p[i].first, d);
        c++;
      }
      i++;
      y++;
    }
    long long prev = x - 1;
    long long nxt = x + y;
    long long t = 0;
    if (prev >= 0) t = a[prev];
    if (nxt < n) t = max(t, a[nxt]);
    long long g = c / k;
    long long e = c % k;
    if (g == 0) {
      if (t < d) {
        cout << "-1";
        return 0;
      } else {
        ans += c * r;
      }
    } else {
      if (u == 1) {
        long long ans1;
        if (t >= d)
          ans += c * r;
        else
          ans += ((c - k) * r + s);
      } else {
        long long ans1 = (e * r + g * s);
        ans += ans1;
      }
    }
  }
  cout << ans << "\n";
}

```