#include <bits/stdc++.h>
using namespace std;
int dx4[4] = {0, -1, 0, 1};
int dy4[4] = {1, 0, -1, 0};
int dx8[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy8[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
int Power(int base, int exp) {
  int ans = 1;
  while (exp > 0) {
    if (exp & 1) ans = (1LL * ans * base) % 1000000007;
    exp = exp >> 1;
    base = (1LL * base * base) % 1000000007;
  }
  return ans;
}
long long int cost = 0;
int n, m, x, k, y;
vector<int> a, b;
void call(int s, int e) {
  if (e - s == 1) return;
  int mx = 0;
  for (int l = s + 1; l < e; l += 1) mx = max(mx, a[l]);
  int lt = (s == -1 ? INT_MIN : a[s]);
  int rt = (e == n ? INT_MIN : a[e]);
  if (e - s - 1 < k && mx > lt && mx > rt) {
    cout << -1 << '\n';
    exit(0);
  }
  long long int tmp = 1000000000000000000;
  if (mx < lt || mx < rt) tmp = min(tmp, 1LL * (e - s - 1) * y);
  for (int t = 1; t < 300000; t += 1) {
    long long int once = e - s - 1 - (1LL * t * k);
    if (once < 0) break;
    tmp = min(tmp, 1LL * t * x + once * y);
  }
  cost += tmp;
}
void input() {
  cin >> n >> m >> x >> k >> y;
  a.resize(n);
  b.resize(m);
  for (int i = 0; i < n; i += 1) cin >> a[i];
  for (int i = 0; i < m; i += 1) cin >> b[i];
  vector<int> pos;
  pos.push_back(-1);
  int i = 0, j = 0;
  while (i < n && j < m) {
    if (a[i] == b[j]) {
      pos.push_back(i);
      j++;
    }
    i++;
  }
  pos.push_back(n);
  if (pos.size() != m + 2) {
    cout << -1 << '\n';
    return;
  }
  for (int i = 0; i < pos.size() - 1; i += 1) call(pos[i], pos[i + 1]);
  cout << cost << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int q = 1;
  while (q--) input();
}

### 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;
int dx4[4] = {0, -1, 0, 1};
int dy4[4] = {1, 0, -1, 0};
int dx8[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy8[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
int Power(int base, int exp) {
  int ans = 1;
  while (exp > 0) {
    if (exp & 1) ans = (1LL * ans * base) % 1000000007;
    exp = exp >> 1;
    base = (1LL * base * base) % 1000000007;
  }
  return ans;
}
long long int cost = 0;
int n, m, x, k, y;
vector<int> a, b;
void call(int s, int e) {
  if (e - s == 1) return;
  int mx = 0;
  for (int l = s + 1; l < e; l += 1) mx = max(mx, a[l]);
  int lt = (s == -1 ? INT_MIN : a[s]);
  int rt = (e == n ? INT_MIN : a[e]);
  if (e - s - 1 < k && mx > lt && mx > rt) {
    cout << -1 << '\n';
    exit(0);
  }
  long long int tmp = 1000000000000000000;
  if (mx < lt || mx < rt) tmp = min(tmp, 1LL * (e - s - 1) * y);
  for (int t = 1; t < 300000; t += 1) {
    long long int once = e - s - 1 - (1LL * t * k);
    if (once < 0) break;
    tmp = min(tmp, 1LL * t * x + once * y);
  }
  cost += tmp;
}
void input() {
  cin >> n >> m >> x >> k >> y;
  a.resize(n);
  b.resize(m);
  for (int i = 0; i < n; i += 1) cin >> a[i];
  for (int i = 0; i < m; i += 1) cin >> b[i];
  vector<int> pos;
  pos.push_back(-1);
  int i = 0, j = 0;
  while (i < n && j < m) {
    if (a[i] == b[j]) {
      pos.push_back(i);
      j++;
    }
    i++;
  }
  pos.push_back(n);
  if (pos.size() != m + 2) {
    cout << -1 << '\n';
    return;
  }
  for (int i = 0; i < pos.size() - 1; i += 1) call(pos[i], pos[i + 1]);
  cout << cost << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int q = 1;
  while (q--) input();
}

```