#include <bits/stdc++.h>
using namespace std;
void read(long long &x) {
  x = 0;
  long long f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    f |= (ch == '-');
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  x = f ? -x : x;
  return;
}
long long n, m, x, y, k;
const long long maxn = 200005;
long long a[maxn], b[maxn], cong[maxn];
long long ans = 0;
bool baoliu[maxn];
long long chong(long long l, long long r, long long sy) {
  if (l > r) return 0;
  long long syy = 0;
  for (long long i = l; i <= r; i++) {
    syy = max(syy, a[i]);
  }
  if (syy > sy) {
    if (r - l + 1 < k) return -1;
    if (k * y > x) {
      return (r - l + 1) / k * x + (r - l + 1) % k * y;
    } else {
      return x + (r - l + 1 - k) * y;
    }
  }
  if (k * y > x) {
    return (r - l + 1) / k * x + (r - l + 1) % k * y;
  } else {
    return (r - l + 1) * y;
  }
}
signed main() {
  read(n), read(m), read(x), read(k), read(y);
  for (long long i = 1; i <= n; ++i) read(a[i]), cong[a[i]] = i;
  long long cnt = 1;
  for (long long i = 1; i <= m; ++i) {
    read(b[i]);
    while (a[cnt] != b[i] && cnt <= n) {
      cnt++;
    }
    if (cnt == n + 1) {
      printf("-1");
      return 0;
    }
    baoliu[cnt] = 1;
  }
  long long las = 0;
  for (long long i = 1; i <= n; ++i) {
    if (baoliu[i]) {
      long long haa = chong(las + 1, i - 1, max(a[las], a[i]));
      if (haa == -1) {
        printf("-1");
        return 0;
      }
      ans += haa;
      las = i;
    }
  }
  long long haa = chong(las + 1, n, a[las]);
  if (haa == -1) {
    printf("-1");
    return 0;
  }
  ans += haa;
  printf("%lld\n", ans);
}

### 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;
void read(long long &x) {
  x = 0;
  long long f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    f |= (ch == '-');
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  x = f ? -x : x;
  return;
}
long long n, m, x, y, k;
const long long maxn = 200005;
long long a[maxn], b[maxn], cong[maxn];
long long ans = 0;
bool baoliu[maxn];
long long chong(long long l, long long r, long long sy) {
  if (l > r) return 0;
  long long syy = 0;
  for (long long i = l; i <= r; i++) {
    syy = max(syy, a[i]);
  }
  if (syy > sy) {
    if (r - l + 1 < k) return -1;
    if (k * y > x) {
      return (r - l + 1) / k * x + (r - l + 1) % k * y;
    } else {
      return x + (r - l + 1 - k) * y;
    }
  }
  if (k * y > x) {
    return (r - l + 1) / k * x + (r - l + 1) % k * y;
  } else {
    return (r - l + 1) * y;
  }
}
signed main() {
  read(n), read(m), read(x), read(k), read(y);
  for (long long i = 1; i <= n; ++i) read(a[i]), cong[a[i]] = i;
  long long cnt = 1;
  for (long long i = 1; i <= m; ++i) {
    read(b[i]);
    while (a[cnt] != b[i] && cnt <= n) {
      cnt++;
    }
    if (cnt == n + 1) {
      printf("-1");
      return 0;
    }
    baoliu[cnt] = 1;
  }
  long long las = 0;
  for (long long i = 1; i <= n; ++i) {
    if (baoliu[i]) {
      long long haa = chong(las + 1, i - 1, max(a[las], a[i]));
      if (haa == -1) {
        printf("-1");
        return 0;
      }
      ans += haa;
      las = i;
    }
  }
  long long haa = chong(las + 1, n, a[las]);
  if (haa == -1) {
    printf("-1");
    return 0;
  }
  ans += haa;
  printf("%lld\n", ans);
}

```