#include <bits/stdc++.h>
using namespace std;
const int maxx = 600010;
long long t, n, m, x, k, y;
long long ans = 0, total = 0, sum = 0, cnt = 0, l = 0, r = 0, maxn = 0;
long long a[maxx], b[maxx], l1[maxx], kl[maxx];
int main() {
  scanf("%lld%lld", &n, &m);
  scanf("%lld%lld%lld", &x, &k, &y);
  for (register int i = 1; i <= n; i++) scanf("%lld", &a[i]), l1[a[i]] = i;
  for (register int i = 1; i <= m; i++) scanf("%lld", &b[i]);
  for (register int i = 1; i < m; i++)
    if (l1[b[i]] < l1[b[i - 1]]) {
      printf("-1");
      return 0;
    }
  b[++m] = 0;
  l1[b[m]] = n + 1;
  b[0] = n + 1;
  l1[b[n + 1]] = n + 1;
  for (register int i = m - 1; i >= 0; i--) {
    maxn = 0;
    for (register int j = l1[b[i + 1]] - 1; j >= l1[b[i]] + 1; j--)
      maxn = max(maxn, a[j]);
    kl[i] = maxn;
  }
  for (register int i = m - 1; i >= 0; i--) {
    sum = l1[b[i + 1]] - l1[b[i]] - 1;
    if (!sum) continue;
    if (sum == 1) {
      if (k == 1) {
        if (a[l1[b[i + 1]] - 1] > a[l1[b[i + 1]]] &&
            a[l1[b[i + 1]] - 1] > a[l1[b[i]]])
          ans += x;
        else {
          if (y <= x)
            ans += y;
          else
            ans += x;
        }
      } else {
        if (a[l1[b[i + 1]] - 1] > a[l1[b[i + 1]]] &&
            a[l1[b[i + 1]] - 1] > a[l1[b[i]]]) {
          printf("-1");
          return 0;
        } else
          ans += y;
      }
    } else {
      if (kl[i] > a[l1[b[i + 1]]] && kl[i] > a[l1[b[i]]]) {
        if (sum < k) {
          printf("-1");
          return 0;
        } else {
          if (y * k >= x) {
            ans += sum / k * x;
            sum %= k;
            ans += y * sum;
          } else {
            sum -= k;
            ans += x + sum * y;
          }
        }
      } else {
        if (sum < k)
          ans += y * sum;
        else {
          if (y * k >= x) {
            ans += sum / k * x;
            sum %= k;
            ans += y * sum;
          } else
            ans += y * sum;
        }
      }
    }
  }
  printf("%lld", ans);
  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 maxx = 600010;
long long t, n, m, x, k, y;
long long ans = 0, total = 0, sum = 0, cnt = 0, l = 0, r = 0, maxn = 0;
long long a[maxx], b[maxx], l1[maxx], kl[maxx];
int main() {
  scanf("%lld%lld", &n, &m);
  scanf("%lld%lld%lld", &x, &k, &y);
  for (register int i = 1; i <= n; i++) scanf("%lld", &a[i]), l1[a[i]] = i;
  for (register int i = 1; i <= m; i++) scanf("%lld", &b[i]);
  for (register int i = 1; i < m; i++)
    if (l1[b[i]] < l1[b[i - 1]]) {
      printf("-1");
      return 0;
    }
  b[++m] = 0;
  l1[b[m]] = n + 1;
  b[0] = n + 1;
  l1[b[n + 1]] = n + 1;
  for (register int i = m - 1; i >= 0; i--) {
    maxn = 0;
    for (register int j = l1[b[i + 1]] - 1; j >= l1[b[i]] + 1; j--)
      maxn = max(maxn, a[j]);
    kl[i] = maxn;
  }
  for (register int i = m - 1; i >= 0; i--) {
    sum = l1[b[i + 1]] - l1[b[i]] - 1;
    if (!sum) continue;
    if (sum == 1) {
      if (k == 1) {
        if (a[l1[b[i + 1]] - 1] > a[l1[b[i + 1]]] &&
            a[l1[b[i + 1]] - 1] > a[l1[b[i]]])
          ans += x;
        else {
          if (y <= x)
            ans += y;
          else
            ans += x;
        }
      } else {
        if (a[l1[b[i + 1]] - 1] > a[l1[b[i + 1]]] &&
            a[l1[b[i + 1]] - 1] > a[l1[b[i]]]) {
          printf("-1");
          return 0;
        } else
          ans += y;
      }
    } else {
      if (kl[i] > a[l1[b[i + 1]]] && kl[i] > a[l1[b[i]]]) {
        if (sum < k) {
          printf("-1");
          return 0;
        } else {
          if (y * k >= x) {
            ans += sum / k * x;
            sum %= k;
            ans += y * sum;
          } else {
            sum -= k;
            ans += x + sum * y;
          }
        }
      } else {
        if (sum < k)
          ans += y * sum;
        else {
          if (y * k >= x) {
            ans += sum / k * x;
            sum %= k;
            ans += y * sum;
          } else
            ans += y * sum;
        }
      }
    }
  }
  printf("%lld", ans);
  return 0;
}

```