#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10;
int read() {
  int ans = 0;
  bool f = 0;
  char ch = getchar();
  while (ch < '0' or ch > '9') {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch >= '0' and ch <= '9') {
    ans = (ans << 1) + (ans << 3) + (ch ^ 48);
    ch = getchar();
  }
  return f ? ~ans + 1 : ans;
}
int n, m, x, y, k;
int a[maxn], b[maxn], mmp[maxn];
long long ans;
vector<int> g;
int main() {
  n = read(), m = read(), x = read(), k = read(), y = read();
  for (int i = 1; i <= n; ++i) a[i] = read(), mmp[a[i]] = i;
  for (int i = 1; i <= m; ++i) b[i] = read(), g.push_back(mmp[b[i]]);
  g.push_back(n + 1);
  bool flag = 0;
  for (int i = 2; i <= m; ++i)
    if (mmp[b[i - 1]] > mmp[b[i]]) {
      flag = 1;
      break;
    }
  if (flag) return printf("-1"), 0;
  int last = 0;
  for (int i = 0; i < g.size(); ++i) {
    int now = g[i], maxx = 0, len = now - last - 1;
    for (int j = last + 1; j < now; ++j) maxx = max(maxx, a[j]);
    if (maxx > a[last] and maxx > a[now]) {
      if (len < k) {
        flag = 1;
        break;
      }
      ans += min(1ll * x + 1ll * (len - k) * y,
                 1ll * (len / k) * x + 1ll * (len % k) * y);
      last = now;
      continue;
    }
    ans += min(1ll * len * y, 1ll * (len / k) * x + 1ll * (len % k) * y);
    last = now;
  }
  if (flag) return printf("-1"), 0;
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 maxn = 3e5 + 10;
int read() {
  int ans = 0;
  bool f = 0;
  char ch = getchar();
  while (ch < '0' or ch > '9') {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch >= '0' and ch <= '9') {
    ans = (ans << 1) + (ans << 3) + (ch ^ 48);
    ch = getchar();
  }
  return f ? ~ans + 1 : ans;
}
int n, m, x, y, k;
int a[maxn], b[maxn], mmp[maxn];
long long ans;
vector<int> g;
int main() {
  n = read(), m = read(), x = read(), k = read(), y = read();
  for (int i = 1; i <= n; ++i) a[i] = read(), mmp[a[i]] = i;
  for (int i = 1; i <= m; ++i) b[i] = read(), g.push_back(mmp[b[i]]);
  g.push_back(n + 1);
  bool flag = 0;
  for (int i = 2; i <= m; ++i)
    if (mmp[b[i - 1]] > mmp[b[i]]) {
      flag = 1;
      break;
    }
  if (flag) return printf("-1"), 0;
  int last = 0;
  for (int i = 0; i < g.size(); ++i) {
    int now = g[i], maxx = 0, len = now - last - 1;
    for (int j = last + 1; j < now; ++j) maxx = max(maxx, a[j]);
    if (maxx > a[last] and maxx > a[now]) {
      if (len < k) {
        flag = 1;
        break;
      }
      ans += min(1ll * x + 1ll * (len - k) * y,
                 1ll * (len / k) * x + 1ll * (len % k) * y);
      last = now;
      continue;
    }
    ans += min(1ll * len * y, 1ll * (len / k) * x + 1ll * (len % k) * y);
    last = now;
  }
  if (flag) return printf("-1"), 0;
  cout << ans << endl;
  return 0;
}

```