#include <bits/stdc++.h>
using namespace std;
long long n, m, x, k, y, maks, tmp, a[200005], b[200005], pos[200005],
    dp[2][200005], ans;
bool f;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> x >> k >> y;
  for (int i = 1; i <= n; i++) {
    dp[0][i] = i * y;
    if (i >= k) {
      if (x <= k * y) {
        maks = i / k;
        dp[1][i] = maks * x + (i - maks * k) * y;
      } else {
        dp[1][i] = x + (i - k) * y;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  for (int i = 0; i < m; i++) {
    if (pos[b[i]] < pos[b[i + 1]]) {
      tmp = pos[b[i + 1]] - pos[b[i]] - 1, maks = 0;
      for (int j = pos[b[i]] + 1; j < pos[b[i + 1]]; j++) {
        maks = max(a[j], maks);
      }
      if (tmp < k) {
        if (maks > max(b[i], b[i + 1])) {
          f = true;
          break;
        }
        ans += dp[0][tmp];
      } else {
        if (maks > max(b[i], b[i + 1])) {
          ans += dp[1][tmp];
        } else {
          ans += min(dp[1][tmp], dp[0][tmp]);
        }
      }
    } else {
      f = true;
      break;
    }
  }
  maks = 0, tmp = n - pos[b[m]];
  for (int i = pos[b[m]] + 1; i <= n; i++) {
    maks = max(maks, a[i]);
  }
  if (tmp < k) {
    if (maks > b[m]) {
      f = true;
    }
    ans += dp[0][tmp];
  } else {
    if (maks > b[m]) {
      ans += dp[1][tmp];
    } else {
      ans += min(dp[0][tmp], dp[1][tmp]);
    }
  }
  if (f) {
    cout << -1 << '\n';
  } else {
    cout << ans << '\n';
  }
}

### Prompt

Generate 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;
long long n, m, x, k, y, maks, tmp, a[200005], b[200005], pos[200005],
    dp[2][200005], ans;
bool f;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> x >> k >> y;
  for (int i = 1; i <= n; i++) {
    dp[0][i] = i * y;
    if (i >= k) {
      if (x <= k * y) {
        maks = i / k;
        dp[1][i] = maks * x + (i - maks * k) * y;
      } else {
        dp[1][i] = x + (i - k) * y;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  for (int i = 0; i < m; i++) {
    if (pos[b[i]] < pos[b[i + 1]]) {
      tmp = pos[b[i + 1]] - pos[b[i]] - 1, maks = 0;
      for (int j = pos[b[i]] + 1; j < pos[b[i + 1]]; j++) {
        maks = max(a[j], maks);
      }
      if (tmp < k) {
        if (maks > max(b[i], b[i + 1])) {
          f = true;
          break;
        }
        ans += dp[0][tmp];
      } else {
        if (maks > max(b[i], b[i + 1])) {
          ans += dp[1][tmp];
        } else {
          ans += min(dp[1][tmp], dp[0][tmp]);
        }
      }
    } else {
      f = true;
      break;
    }
  }
  maks = 0, tmp = n - pos[b[m]];
  for (int i = pos[b[m]] + 1; i <= n; i++) {
    maks = max(maks, a[i]);
  }
  if (tmp < k) {
    if (maks > b[m]) {
      f = true;
    }
    ans += dp[0][tmp];
  } else {
    if (maks > b[m]) {
      ans += dp[1][tmp];
    } else {
      ans += min(dp[0][tmp], dp[1][tmp]);
    }
  }
  if (f) {
    cout << -1 << '\n';
  } else {
    cout << ans << '\n';
  }
}

```