#include <bits/stdc++.h>
#pragma GCC optimize(2)
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int t, n, m, k, x, y, a[N], b[N], L, R, pos;
long long ans;
bool check() {
  int mx = max(a[L - 1], a[R]), mxcnt = 0, len = R - L;
  for (int i = L; i < R; ++i)
    if (a[i] > mx) ++mxcnt;
  if (len < k) {
    if (mxcnt) return 1;
    ans += 1ll * y * len;
  } else {
    if (mxcnt) len -= k, ans += x;
    ans += min(1ll * y * len, 1ll * (len / k) * x + 1ll * (len % k) * y);
  }
  return 0;
}
long long solve() {
  cin >> n >> m >> x >> k >> y;
  a[n] = -1;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) cin >> b[i];
  for (R = 0; R < n; ++R) {
    if (a[R] == b[pos]) {
      if (L != R - 1 || L + R != 0)
        if (check()) return -1;
      ++pos, L = R + 1;
    }
  }
  if (pos != m) return -1;
  if (L < n)
    if (check()) return -1;
  return ans;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cout << solve();
  return 0;
}

### 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>
#pragma GCC optimize(2)
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int t, n, m, k, x, y, a[N], b[N], L, R, pos;
long long ans;
bool check() {
  int mx = max(a[L - 1], a[R]), mxcnt = 0, len = R - L;
  for (int i = L; i < R; ++i)
    if (a[i] > mx) ++mxcnt;
  if (len < k) {
    if (mxcnt) return 1;
    ans += 1ll * y * len;
  } else {
    if (mxcnt) len -= k, ans += x;
    ans += min(1ll * y * len, 1ll * (len / k) * x + 1ll * (len % k) * y);
  }
  return 0;
}
long long solve() {
  cin >> n >> m >> x >> k >> y;
  a[n] = -1;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) cin >> b[i];
  for (R = 0; R < n; ++R) {
    if (a[R] == b[pos]) {
      if (L != R - 1 || L + R != 0)
        if (check()) return -1;
      ++pos, L = R + 1;
    }
  }
  if (pos != m) return -1;
  if (L < n)
    if (check()) return -1;
  return ans;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cout << solve();
  return 0;
}

```