#include <bits/stdc++.h>
using namespace std;
int N, M, X, K, Y;
vector<int> a, b;
long long myfunc(int l, int r) {
  int num = r - l - 1;
  if (!num) return 0LL;
  int tmp = *max_element(a.begin() + l + 1, a.begin() + r);
  bool ok = 0;
  if (a[l] > tmp || a[r] > tmp) ok = 1;
  if (num >= K) {
    if (1LL * X <= 1LL * Y * K) {
      return 1LL * num / K * X + 1LL * num % K * Y;
    } else {
      if (ok) {
        return 1LL * num * Y;
      } else {
        return 1LL * X + 1LL * (num - K) * Y;
      }
    }
  } else {
    if (ok)
      return 1LL * Y * num;
    else
      return -1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M >> X >> K >> Y;
  a = vector<int>(N + 2);
  b = vector<int>(M + 2);
  for (int n = 1; n <= N; ++n) cin >> a[n];
  for (int m = 1; m <= M; ++m) cin >> b[m];
  long long ans = 0LL, tmp = 0LL;
  int aidx, bidx = 1;
  int l = 0, r = 1;
  bool ok = 1;
  for (aidx = 1; aidx <= N; ++aidx) {
    if (bidx <= M && a[aidx] == b[bidx]) {
      tmp = myfunc(l, r);
      if (tmp == -1)
        ok = 0;
      else
        ans += tmp;
      bidx++;
      l = aidx;
    }
    r++;
  }
  if (bidx <= M)
    cout << -1;
  else {
    tmp = myfunc(l, r);
    if (tmp == -1)
      ok = 0;
    else
      ans += tmp;
    if (ok)
      cout << ans;
    else
      cout << -1;
  }
  return 0;
}

### Prompt

Your task is to create 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;
vector<int> a, b;
long long myfunc(int l, int r) {
  int num = r - l - 1;
  if (!num) return 0LL;
  int tmp = *max_element(a.begin() + l + 1, a.begin() + r);
  bool ok = 0;
  if (a[l] > tmp || a[r] > tmp) ok = 1;
  if (num >= K) {
    if (1LL * X <= 1LL * Y * K) {
      return 1LL * num / K * X + 1LL * num % K * Y;
    } else {
      if (ok) {
        return 1LL * num * Y;
      } else {
        return 1LL * X + 1LL * (num - K) * Y;
      }
    }
  } else {
    if (ok)
      return 1LL * Y * num;
    else
      return -1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M >> X >> K >> Y;
  a = vector<int>(N + 2);
  b = vector<int>(M + 2);
  for (int n = 1; n <= N; ++n) cin >> a[n];
  for (int m = 1; m <= M; ++m) cin >> b[m];
  long long ans = 0LL, tmp = 0LL;
  int aidx, bidx = 1;
  int l = 0, r = 1;
  bool ok = 1;
  for (aidx = 1; aidx <= N; ++aidx) {
    if (bidx <= M && a[aidx] == b[bidx]) {
      tmp = myfunc(l, r);
      if (tmp == -1)
        ok = 0;
      else
        ans += tmp;
      bidx++;
      l = aidx;
    }
    r++;
  }
  if (bidx <= M)
    cout << -1;
  else {
    tmp = myfunc(l, r);
    if (tmp == -1)
      ok = 0;
    else
      ans += tmp;
    if (ok)
      cout << ans;
    else
      cout << -1;
  }
  return 0;
}

```