#include <bits/stdc++.h>
using namespace std;
bool subsequence(vector<long long int> &v1, vector<long long int> &v2,
                 vector<long long int> &v3, long long int n, long long int m) {
  long long int i = 0, j = 0;
  while (1) {
    if (v1[i] == v2[j]) {
      v3.push_back(i);
      i++;
      j++;
    } else
      i++;
    if (j == m || i == n) break;
  }
  if (j == m) return true;
  if (i == n) return false;
}
int main() {
  long long int n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  vector<long long int> v1(n), v2(m);
  for (auto &it : v1) cin >> it;
  for (auto &it : v2) cin >> it;
  vector<long long int> v3;
  bool b1 = subsequence(v1, v2, v3, n, m);
  if (b1 == false || m > n)
    cout << "-1";
  else {
    long long int ans = 0;
    double t1 = double(x) / k;
    long long int t2 = y;
    long long int i = 0;
    v3.push_back(n);
    v3.insert(v3.begin(), -1);
    v2.push_back(0);
    v2.insert(v2.begin(), 0);
    while (1) {
      long long int start = v3[i];
      long long int end = v3[i + 1];
      if (end == start + 1)
        ans += 0;
      else if (t1 < t2) {
        long long int a11 =
            *max_element(v1.begin() + start + 1, v1.begin() + end);
        if (k > end - start - 1 && a11 > max(v2[i], v2[i + 1])) {
          ans = -1;
          break;
        } else
          ans += ((end - start - 1) / k) * x + ((end - start - 1) % k) * t2;
      } else if (t1 == t2) {
        ans += (end - start - 1) * t2;
      } else {
        long long int a22 =
            *max_element(v1.begin() + start + 1, v1.begin() + end);
        if (k > end - start - 1 && a22 > max(v2[i], v2[i + 1])) {
          ans = -1;
          break;
        } else if (a22 > max(v2[i], v2[i + 1]))
          ans += (x + (end - start - 1 - k) * t2);
        else
          ans += (end - start - 1) * t2;
      }
      i++;
      if (i == m + 1) break;
    }
    cout << ans;
  }
}

### Prompt

Create a solution in Cpp for 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;
bool subsequence(vector<long long int> &v1, vector<long long int> &v2,
                 vector<long long int> &v3, long long int n, long long int m) {
  long long int i = 0, j = 0;
  while (1) {
    if (v1[i] == v2[j]) {
      v3.push_back(i);
      i++;
      j++;
    } else
      i++;
    if (j == m || i == n) break;
  }
  if (j == m) return true;
  if (i == n) return false;
}
int main() {
  long long int n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  vector<long long int> v1(n), v2(m);
  for (auto &it : v1) cin >> it;
  for (auto &it : v2) cin >> it;
  vector<long long int> v3;
  bool b1 = subsequence(v1, v2, v3, n, m);
  if (b1 == false || m > n)
    cout << "-1";
  else {
    long long int ans = 0;
    double t1 = double(x) / k;
    long long int t2 = y;
    long long int i = 0;
    v3.push_back(n);
    v3.insert(v3.begin(), -1);
    v2.push_back(0);
    v2.insert(v2.begin(), 0);
    while (1) {
      long long int start = v3[i];
      long long int end = v3[i + 1];
      if (end == start + 1)
        ans += 0;
      else if (t1 < t2) {
        long long int a11 =
            *max_element(v1.begin() + start + 1, v1.begin() + end);
        if (k > end - start - 1 && a11 > max(v2[i], v2[i + 1])) {
          ans = -1;
          break;
        } else
          ans += ((end - start - 1) / k) * x + ((end - start - 1) % k) * t2;
      } else if (t1 == t2) {
        ans += (end - start - 1) * t2;
      } else {
        long long int a22 =
            *max_element(v1.begin() + start + 1, v1.begin() + end);
        if (k > end - start - 1 && a22 > max(v2[i], v2[i + 1])) {
          ans = -1;
          break;
        } else if (a22 > max(v2[i], v2[i + 1]))
          ans += (x + (end - start - 1 - k) * t2);
        else
          ans += (end - start - 1) * t2;
      }
      i++;
      if (i == m + 1) break;
    }
    cout << ans;
  }
}

```