#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int lim = 200010;
  long long int n, m, a1[lim], a2[lim], ind[lim], ma[lim], ma2[lim], x, y, k, c,
      mx, tmp;
  long long int cnt;
  bool flg;
  string st;
  cin >> n >> m;
  c = 0;
  cin >> x >> k >> y;
  ind[c] = -1;
  for (long long int i = 0; i < n; i++) cin >> a1[i];
  for (long long int i = 0; i < m; i++) cin >> a2[i];
  for (long long int i = 0; i < n; i++)
    if (a1[i] == a2[c]) {
      ++c;
      ind[c] = i;
    }
  if (c == m && m == n)
    cout << 0;
  else if (c != m)
    cout << -1;
  else {
    ind[++c] = n;
    c++;
    cnt = 0;
    flg = true;
    for (long long int i = 0; i < c - 1; i++) {
      mx = 0;
      for (long long int j = ind[i] + 1; j < ind[i + 1]; j++)
        if (a1[j] > mx) mx = a1[j];
      ma[i] = mx;
    }
    ma2[0] = a1[ind[1]];
    for (long long int i = 1; i < c - 2; i++) {
      if (a1[ind[i]] > a1[ind[i + 1]])
        mx = a1[ind[i]];
      else
        mx = a1[ind[i + 1]];
      ma2[i] = mx;
    }
    ma2[c - 2] = a1[ind[c - 2]];
    if ((double)x / k < y) {
      for (long long int i = 0; i < c - 1; i++) {
        tmp = ind[i + 1] - ind[i] - 1;
        if (tmp < k && tmp != 0) {
          if (ma[i] < ma2[i]) {
            cnt += y * tmp;
          } else {
            flg = false;
            break;
          }
        } else {
          cnt += (tmp / k) * x + (tmp % k) * y;
        }
      }
    } else {
      for (long long int i = 0; i < c - 1; i++) {
        tmp = ind[i + 1] - ind[i] - 1;
        if (tmp != 0) {
          if (ma[i] < ma2[i]) {
            cnt += y * tmp;
          } else if (tmp >= k) {
            cnt += x;
            tmp -= k;
            cnt += y * tmp;
          } else {
            flg = false;
            break;
          }
        }
      }
    }
    if (flg)
      cout << cnt;
    else
      cout << -1;
    cout << "\n";
  }
  return 0;
}

### 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;
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int lim = 200010;
  long long int n, m, a1[lim], a2[lim], ind[lim], ma[lim], ma2[lim], x, y, k, c,
      mx, tmp;
  long long int cnt;
  bool flg;
  string st;
  cin >> n >> m;
  c = 0;
  cin >> x >> k >> y;
  ind[c] = -1;
  for (long long int i = 0; i < n; i++) cin >> a1[i];
  for (long long int i = 0; i < m; i++) cin >> a2[i];
  for (long long int i = 0; i < n; i++)
    if (a1[i] == a2[c]) {
      ++c;
      ind[c] = i;
    }
  if (c == m && m == n)
    cout << 0;
  else if (c != m)
    cout << -1;
  else {
    ind[++c] = n;
    c++;
    cnt = 0;
    flg = true;
    for (long long int i = 0; i < c - 1; i++) {
      mx = 0;
      for (long long int j = ind[i] + 1; j < ind[i + 1]; j++)
        if (a1[j] > mx) mx = a1[j];
      ma[i] = mx;
    }
    ma2[0] = a1[ind[1]];
    for (long long int i = 1; i < c - 2; i++) {
      if (a1[ind[i]] > a1[ind[i + 1]])
        mx = a1[ind[i]];
      else
        mx = a1[ind[i + 1]];
      ma2[i] = mx;
    }
    ma2[c - 2] = a1[ind[c - 2]];
    if ((double)x / k < y) {
      for (long long int i = 0; i < c - 1; i++) {
        tmp = ind[i + 1] - ind[i] - 1;
        if (tmp < k && tmp != 0) {
          if (ma[i] < ma2[i]) {
            cnt += y * tmp;
          } else {
            flg = false;
            break;
          }
        } else {
          cnt += (tmp / k) * x + (tmp % k) * y;
        }
      }
    } else {
      for (long long int i = 0; i < c - 1; i++) {
        tmp = ind[i + 1] - ind[i] - 1;
        if (tmp != 0) {
          if (ma[i] < ma2[i]) {
            cnt += y * tmp;
          } else if (tmp >= k) {
            cnt += x;
            tmp -= k;
            cnt += y * tmp;
          } else {
            flg = false;
            break;
          }
        }
      }
    }
    if (flg)
      cout << cnt;
    else
      cout << -1;
    cout << "\n";
  }
  return 0;
}

```