#include <bits/stdc++.h>
using namespace std;
int n, m, x, k, y, maxx = 0, tot = 0;
int const N = 2e5 + 10;
int a[N], b[N];
long long ans = 0;
struct P {
  int ll, rr, mm;
} num[N];
int main() {
  memset(a, 0, sizeof(a));
  scanf("%d%d", &n, &m);
  scanf("%d%d%d", &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]);
  memset(num, 0, sizeof(num));
  int l = 1, r = 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] == b[r]) {
      if (i - 1 >= l) {
        num[++tot].ll = l;
        num[tot].rr = i - 1;
        num[tot].mm = maxx;
      }
      r++;
      l = i + 1;
      maxx = 0;
    } else
      maxx = max(maxx, a[i]);
  }
  if (n >= l) {
    num[++tot].ll = l;
    num[tot].rr = n;
    num[tot].mm = maxx;
  }
  if (r <= m) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= tot; i++) {
    l = num[i].ll, r = num[i].rr, maxx = num[i].mm;
    int len = r - l + 1;
    int maxx2 = max(a[l - 1], a[r + 1]);
    if (maxx2 <= maxx && len < k) {
      printf("-1\n");
      return 0;
    }
    if (x >= k * y && maxx2 > maxx)
      ans += 1ll * len * y;
    else if (x >= k * y && r - l + 1 >= k)
      ans += 1ll * (len - k) * y + x;
    else
      ans += 1ll * (len / k) * x + 1ll * y * (len % k);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Generate 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;
int n, m, x, k, y, maxx = 0, tot = 0;
int const N = 2e5 + 10;
int a[N], b[N];
long long ans = 0;
struct P {
  int ll, rr, mm;
} num[N];
int main() {
  memset(a, 0, sizeof(a));
  scanf("%d%d", &n, &m);
  scanf("%d%d%d", &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]);
  memset(num, 0, sizeof(num));
  int l = 1, r = 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] == b[r]) {
      if (i - 1 >= l) {
        num[++tot].ll = l;
        num[tot].rr = i - 1;
        num[tot].mm = maxx;
      }
      r++;
      l = i + 1;
      maxx = 0;
    } else
      maxx = max(maxx, a[i]);
  }
  if (n >= l) {
    num[++tot].ll = l;
    num[tot].rr = n;
    num[tot].mm = maxx;
  }
  if (r <= m) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= tot; i++) {
    l = num[i].ll, r = num[i].rr, maxx = num[i].mm;
    int len = r - l + 1;
    int maxx2 = max(a[l - 1], a[r + 1]);
    if (maxx2 <= maxx && len < k) {
      printf("-1\n");
      return 0;
    }
    if (x >= k * y && maxx2 > maxx)
      ans += 1ll * len * y;
    else if (x >= k * y && r - l + 1 >= k)
      ans += 1ll * (len - k) * y + x;
    else
      ans += 1ll * (len / k) * x + 1ll * y * (len % k);
  }
  printf("%lld\n", ans);
  return 0;
}

```