#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int a[maxn], b[maxn];
map<int, int> mp;
vector<int> vec;
long long x, y;
int n, m, k;
long long sum;
bool solve(int i, int j) {
  if (j - i == 1 || ((i == 0) && (j == n + 1))) return true;
  int P, l, r, L, R, len;
  P = 0, l = i + 1;
  r = j - 1;
  L = a[i];
  R = a[j];
  len = r - l + 1;
  for (int t = l; t <= r; t++) {
    P = max(P, a[t]);
  }
  int tim = len % k;
  if (tim) {
    sum += y * tim;
    len = len / k * k;
    if (len == 0) {
      if (P > L && P > R)
        return false;
      else
        return true;
    }
  }
  if (x <= y * k) {
    sum += x * len / k;
  } else {
    if (P < L || P < R) {
      sum += y * len;
    } else {
      sum += x + y * (len - k);
    }
  }
  return true;
}
int main() {
  scanf("%d%d%lld%d%lld", &n, &m, &x, &k, &y);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    mp[a[i]] = i;
  }
  vec.push_back(0);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &b[i]);
    vec.push_back(mp[b[i]]);
  }
  vec.push_back(n + 1);
  for (int i = 1; i < vec.size(); i++) {
    if (vec[i] < vec[i - 1]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 0; i < vec.size() - 1; i++) {
    if (!solve(vec[i], vec[i + 1])) {
      puts("-1");
      return 0;
    }
  }
  printf("%lld\n", sum);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 = 2e5 + 10;
int a[maxn], b[maxn];
map<int, int> mp;
vector<int> vec;
long long x, y;
int n, m, k;
long long sum;
bool solve(int i, int j) {
  if (j - i == 1 || ((i == 0) && (j == n + 1))) return true;
  int P, l, r, L, R, len;
  P = 0, l = i + 1;
  r = j - 1;
  L = a[i];
  R = a[j];
  len = r - l + 1;
  for (int t = l; t <= r; t++) {
    P = max(P, a[t]);
  }
  int tim = len % k;
  if (tim) {
    sum += y * tim;
    len = len / k * k;
    if (len == 0) {
      if (P > L && P > R)
        return false;
      else
        return true;
    }
  }
  if (x <= y * k) {
    sum += x * len / k;
  } else {
    if (P < L || P < R) {
      sum += y * len;
    } else {
      sum += x + y * (len - k);
    }
  }
  return true;
}
int main() {
  scanf("%d%d%lld%d%lld", &n, &m, &x, &k, &y);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    mp[a[i]] = i;
  }
  vec.push_back(0);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &b[i]);
    vec.push_back(mp[b[i]]);
  }
  vec.push_back(n + 1);
  for (int i = 1; i < vec.size(); i++) {
    if (vec[i] < vec[i - 1]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 0; i < vec.size() - 1; i++) {
    if (!solve(vec[i], vec[i + 1])) {
      puts("-1");
      return 0;
    }
  }
  printf("%lld\n", sum);
  return 0;
}

```