#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 10;
const long long inf = 1e18;
long long a[N], b[N];
signed main() {
  long long n, m;
  scanf("%lld%lld", &n, &m);
  long long x, k, y;
  scanf("%lld%lld%lld", &x, &k, &y);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1; i <= m; i++) scanf("%lld", &b[i]);
  a[n + 1] = b[n + 1] = 0;
  long long now = 1, flag = 0, ans = 0;
  for (long long i = 1; i <= m + 1; i++) {
    long long len = 0, maxn = -1;
    while (now <= n && a[now] != b[i]) {
      maxn = max(maxn, a[now]);
      now++, len++;
    }
    if (now == n + 2) {
      flag = 1;
      break;
    }
    now++;
    long long res = inf;
    for (long long j = 1; j * k <= len; j++) {
      res = min(res, j * x + (len - j * k) * y);
    }
    if (maxn < b[i] || (maxn < b[i - 1] && i - 1 > 0)) res = min(res, len * y);
    if (res == inf) {
      flag = 1;
      break;
    }
    ans += res;
  }
  if (flag)
    puts("-1");
  else
    printf("%lld\n", ans);
}

### 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 long long N = 2e5 + 10;
const long long inf = 1e18;
long long a[N], b[N];
signed main() {
  long long n, m;
  scanf("%lld%lld", &n, &m);
  long long x, k, y;
  scanf("%lld%lld%lld", &x, &k, &y);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1; i <= m; i++) scanf("%lld", &b[i]);
  a[n + 1] = b[n + 1] = 0;
  long long now = 1, flag = 0, ans = 0;
  for (long long i = 1; i <= m + 1; i++) {
    long long len = 0, maxn = -1;
    while (now <= n && a[now] != b[i]) {
      maxn = max(maxn, a[now]);
      now++, len++;
    }
    if (now == n + 2) {
      flag = 1;
      break;
    }
    now++;
    long long res = inf;
    for (long long j = 1; j * k <= len; j++) {
      res = min(res, j * x + (len - j * k) * y);
    }
    if (maxn < b[i] || (maxn < b[i - 1] && i - 1 > 0)) res = min(res, len * y);
    if (res == inf) {
      flag = 1;
      break;
    }
    ans += res;
  }
  if (flag)
    puts("-1");
  else
    printf("%lld\n", ans);
}

```