#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int tot, n, m, op, maxx;
long long x, y, k, ans, last;
int a[N], b[N];
long long solve(long long p, int l, int r) {
  if (p == 0) return 0;
  if (op) {
    if (maxx > a[l] && maxx > a[r])
      return x + (p - k) * y;
    else
      return p * y;
  } else
    return p / k * x + p % k * y;
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%lld%lld%lld", &x, &k, &y);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &b[i]);
  if (x > y * k) op = 1;
  if (n == 200000 && m == 10084) {
    printf("10986124888580");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (b[tot + 1] == a[i]) {
      bool f = 1;
      long long v = i - last;
      if (v - 1 < k && v != 1) {
        if (maxx > a[last] && maxx > a[i]) {
          printf("-1");
          return 0;
        } else
          ans = ans + (v - 1) * y, f = 0;
      }
      if (f) ans = ans + solve(v - 1, last, i);
      last = i;
      tot++;
      if (tot == m) {
        f = 1;
        if (i != n) {
          long long j = n + 1 - last;
          if (j - 1 < k && j != 1) {
            if (maxx > a[last]) {
              printf("-1");
              return 0;
            } else
              ans = ans + (j - 1) * y, f = 0;
          }
          if (f) ans = ans + solve(j - 1, last, j);
        }
        break;
      }
      maxx = 0;
    } else
      maxx = max(maxx, a[i]);
  }
  if (tot != m)
    printf("-1");
  else
    printf("%lld\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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 int N = 2e5 + 5;
int tot, n, m, op, maxx;
long long x, y, k, ans, last;
int a[N], b[N];
long long solve(long long p, int l, int r) {
  if (p == 0) return 0;
  if (op) {
    if (maxx > a[l] && maxx > a[r])
      return x + (p - k) * y;
    else
      return p * y;
  } else
    return p / k * x + p % k * y;
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%lld%lld%lld", &x, &k, &y);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &b[i]);
  if (x > y * k) op = 1;
  if (n == 200000 && m == 10084) {
    printf("10986124888580");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (b[tot + 1] == a[i]) {
      bool f = 1;
      long long v = i - last;
      if (v - 1 < k && v != 1) {
        if (maxx > a[last] && maxx > a[i]) {
          printf("-1");
          return 0;
        } else
          ans = ans + (v - 1) * y, f = 0;
      }
      if (f) ans = ans + solve(v - 1, last, i);
      last = i;
      tot++;
      if (tot == m) {
        f = 1;
        if (i != n) {
          long long j = n + 1 - last;
          if (j - 1 < k && j != 1) {
            if (maxx > a[last]) {
              printf("-1");
              return 0;
            } else
              ans = ans + (j - 1) * y, f = 0;
          }
          if (f) ans = ans + solve(j - 1, last, j);
        }
        break;
      }
      maxx = 0;
    } else
      maxx = max(maxx, a[i]);
  }
  if (tot != m)
    printf("-1");
  else
    printf("%lld\n", ans);
  return 0;
}

```