#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000005;
const int SQRTN = 1003;
const int LOGN = 22;
const double PI = acos(-1);
const int INF = 1e9;
int MOD = 1000000007;
const int FMOD = 998244353;
const double eps = 1e-9;
long long gcd(long long a, long long b) {
  if (a == 0) return (b);
  return (gcd(b % a, a));
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
void solve() {
  long long n, m1;
  cin >> n >> m1;
  long long x, k, y;
  cin >> x >> k >> y;
  long long arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  long long brr[m1];
  for (int i = 0; i < m1; i++) cin >> brr[i];
  bool flag = false;
  long long i = 0, j = 0, m = -1, pos, l = -1, r;
  long long result = 0;
  while (i < n && j < m1) {
    if (flag == true) break;
    m = max(m, arr[i]);
    if (m == arr[i]) pos = i;
    if (arr[i] != brr[j])
      i++;
    else {
      r = i;
      i++;
      j++;
      int len = r - l - 1;
      if (pos > l && pos < r && len < k) {
        flag = true;
        break;
      } else {
        long long c1, c2;
        if (len % k == 0)
          c1 = x * (len / k);
        else
          c1 = x * (len / k) + (len - k * (len / k)) * y;
        c2 = y * len;
        if (pos > l && pos < r) c2 = y * (len - k) + x;
        result += min(c1, c2);
      }
      l = r;
      m = arr[l];
    }
  }
  if (j == m1 && i < n) {
    m = arr[i - 1];
    pos = i - 1;
    l = i - 1;
    r = n;
    int len = 0;
    while (i < n) {
      m = max(m, arr[i]);
      if (m == arr[i]) pos = i;
      i++;
      len++;
    }
    if (pos > l && pos < r && len < k)
      flag = true;
    else {
      long long c1, c2;
      if (len % k == 0)
        c1 = x * (len / k);
      else
        c1 = x * (len / k) + (len - k * (len / k)) * y;
      c2 = y * len;
      if (pos > l && pos < r) c2 = (len - k) * y + x;
      result += min(c1, c2);
    }
  }
  if (i == n && j < m1) {
    cout << -1 << "\n";
  } else {
    if (flag)
      cout << -1 << "\n";
    else
      cout << result << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  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 MAXN = 1000005;
const int SQRTN = 1003;
const int LOGN = 22;
const double PI = acos(-1);
const int INF = 1e9;
int MOD = 1000000007;
const int FMOD = 998244353;
const double eps = 1e-9;
long long gcd(long long a, long long b) {
  if (a == 0) return (b);
  return (gcd(b % a, a));
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
void solve() {
  long long n, m1;
  cin >> n >> m1;
  long long x, k, y;
  cin >> x >> k >> y;
  long long arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  long long brr[m1];
  for (int i = 0; i < m1; i++) cin >> brr[i];
  bool flag = false;
  long long i = 0, j = 0, m = -1, pos, l = -1, r;
  long long result = 0;
  while (i < n && j < m1) {
    if (flag == true) break;
    m = max(m, arr[i]);
    if (m == arr[i]) pos = i;
    if (arr[i] != brr[j])
      i++;
    else {
      r = i;
      i++;
      j++;
      int len = r - l - 1;
      if (pos > l && pos < r && len < k) {
        flag = true;
        break;
      } else {
        long long c1, c2;
        if (len % k == 0)
          c1 = x * (len / k);
        else
          c1 = x * (len / k) + (len - k * (len / k)) * y;
        c2 = y * len;
        if (pos > l && pos < r) c2 = y * (len - k) + x;
        result += min(c1, c2);
      }
      l = r;
      m = arr[l];
    }
  }
  if (j == m1 && i < n) {
    m = arr[i - 1];
    pos = i - 1;
    l = i - 1;
    r = n;
    int len = 0;
    while (i < n) {
      m = max(m, arr[i]);
      if (m == arr[i]) pos = i;
      i++;
      len++;
    }
    if (pos > l && pos < r && len < k)
      flag = true;
    else {
      long long c1, c2;
      if (len % k == 0)
        c1 = x * (len / k);
      else
        c1 = x * (len / k) + (len - k * (len / k)) * y;
      c2 = y * len;
      if (pos > l && pos < r) c2 = (len - k) * y + x;
      result += min(c1, c2);
    }
  }
  if (i == n && j < m1) {
    cout << -1 << "\n";
  } else {
    if (flag)
      cout << -1 << "\n";
    else
      cout << result << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return (0);
}

```