#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-7;
const double PI = acos(-1);
const int MOD = 998244353;
const long long INF = 1e18;
const int N = 2e5 + 7;
int n, m, k;
long long x, y;
int a[N], b[N], p[N];
inline long long solve() {
  cin >> n >> m >> x >> k >> y;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    p[a[i]] = i;
  }
  for (int i = 1; i <= m; ++i) {
    cin >> b[i];
    if (p[b[i]] == 0 || p[b[i]] < p[b[i - 1]]) return -1;
  }
  long long res = 0;
  b[0] = b[m + 1] = 0;
  for (int i = 0, l, r; i <= m; ++i) {
    l = i == 0 ? 0 : p[b[i]];
    r = i == m ? n + 1 : p[b[i + 1]];
    if (l + 1 == r) continue;
    int mv = 0, len = r - l - 1;
    for (int j = l + 1; j < r; ++j) mv = max(mv, a[j]);
    long long cur = INF;
    if (mv < max(b[i], b[i + 1])) {
      for (int j = 0; j * k <= len; ++j) {
        cur = min(cur, j * x + (len - j * k) * y);
      }
    } else {
      if (len < k) return -1;
      for (int j = 1; j * k <= len; ++j) {
        cur = min(cur, j * x + (len - j * k) * y);
      }
    }
    res += cur;
  }
  return res;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T = 1;
  for (int t = 1; t <= T; ++t) {
    cout << solve() << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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;
const double eps = 1e-7;
const double PI = acos(-1);
const int MOD = 998244353;
const long long INF = 1e18;
const int N = 2e5 + 7;
int n, m, k;
long long x, y;
int a[N], b[N], p[N];
inline long long solve() {
  cin >> n >> m >> x >> k >> y;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    p[a[i]] = i;
  }
  for (int i = 1; i <= m; ++i) {
    cin >> b[i];
    if (p[b[i]] == 0 || p[b[i]] < p[b[i - 1]]) return -1;
  }
  long long res = 0;
  b[0] = b[m + 1] = 0;
  for (int i = 0, l, r; i <= m; ++i) {
    l = i == 0 ? 0 : p[b[i]];
    r = i == m ? n + 1 : p[b[i + 1]];
    if (l + 1 == r) continue;
    int mv = 0, len = r - l - 1;
    for (int j = l + 1; j < r; ++j) mv = max(mv, a[j]);
    long long cur = INF;
    if (mv < max(b[i], b[i + 1])) {
      for (int j = 0; j * k <= len; ++j) {
        cur = min(cur, j * x + (len - j * k) * y);
      }
    } else {
      if (len < k) return -1;
      for (int j = 1; j * k <= len; ++j) {
        cur = min(cur, j * x + (len - j * k) * y);
      }
    }
    res += cur;
  }
  return res;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T = 1;
  for (int t = 1; t <= T; ++t) {
    cout << solve() << endl;
  }
  return 0;
}

```