#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int maxn = 2e5 + 22;
int n, m, k;
int a[maxn], b[maxn], pos[maxn];
vector<int> v;
bool flag = 1;
long long ans, x, y;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%lld%d%lld", &x, &k, &y);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  v.push_back(0);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &b[i]);
    v.push_back(pos[b[i]]);
  }
  v.push_back(n + 1);
  int siz = v.size();
  for (int i = 1; i <= siz - 1; i++) {
    if (v[i] < v[i - 1]) flag = 0;
  }
  if (!flag)
    puts("-1");
  else {
    for (int i = 1; i <= siz - 1; i++) {
      bool f = 1;
      int l = v[i - 1], r = v[i];
      int len = r - l - 1;
      for (int j = l + 1; j <= r - 1; j++) {
        if (a[j] > a[l] && a[j] > a[r]) {
          f = 0;
          break;
        }
      }
      if (f) {
        if (1ll * k * y > x) {
          ans = ans + 1ll * len / k * x + 1ll * len % k * y;
        } else {
          ans = ans + 1ll * len * y;
        }
      } else {
        if (len < k) {
          flag = 0;
          break;
        } else {
          len -= k;
          ans += x;
          if (1ll * k * y > x) {
            ans = ans + 1ll * len / k * x + 1ll * len % k * y;
          } else {
            ans = ans + 1ll * len * y;
          }
        }
      }
    }
    if (!flag)
      puts("-1");
    else
      printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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 int maxn = 2e5 + 22;
int n, m, k;
int a[maxn], b[maxn], pos[maxn];
vector<int> v;
bool flag = 1;
long long ans, x, y;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%lld%d%lld", &x, &k, &y);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  v.push_back(0);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &b[i]);
    v.push_back(pos[b[i]]);
  }
  v.push_back(n + 1);
  int siz = v.size();
  for (int i = 1; i <= siz - 1; i++) {
    if (v[i] < v[i - 1]) flag = 0;
  }
  if (!flag)
    puts("-1");
  else {
    for (int i = 1; i <= siz - 1; i++) {
      bool f = 1;
      int l = v[i - 1], r = v[i];
      int len = r - l - 1;
      for (int j = l + 1; j <= r - 1; j++) {
        if (a[j] > a[l] && a[j] > a[r]) {
          f = 0;
          break;
        }
      }
      if (f) {
        if (1ll * k * y > x) {
          ans = ans + 1ll * len / k * x + 1ll * len % k * y;
        } else {
          ans = ans + 1ll * len * y;
        }
      } else {
        if (len < k) {
          flag = 0;
          break;
        } else {
          len -= k;
          ans += x;
          if (1ll * k * y > x) {
            ans = ans + 1ll * len / k * x + 1ll * len % k * y;
          } else {
            ans = ans + 1ll * len * y;
          }
        }
      }
    }
    if (!flag)
      puts("-1");
    else
      printf("%lld\n", ans);
  }
  return 0;
}

```