#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const int M = 2e3 + 5;
const int INF = 2e9;
const int sz = 15;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1);
template <class T>
inline void read(T &x) {
  char c;
  x = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') x = -1;
  T res = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
  x *= res;
}
int n, m;
long long a, b, c;
char str[N];
int f[N], g[N];
bool vis[N];
vector<int> v;
long long ans;
bool hhh(int l, int r) {
  bool flag = 1;
  for (int i = l + 1; i <= r - 1; i++) {
    if (f[i] > f[l] && f[i] > f[r]) {
      flag = 0;
      break;
    }
  }
  int remain = r - l - 1;
  if (!flag && remain < b) return false;
  ans += (1ll * remain % b) * c;
  remain = remain / b * b;
  if (a < b * c) {
    ans += (1ll * remain / b) * a;
  } else {
    if (flag) {
      ans += 1ll * remain * c;
    } else {
      ans += a;
      ans += 1ll * (remain - b) * c;
    }
  }
  return true;
}
int main() {
  read(n);
  read(m);
  read(a);
  read(b);
  read(c);
  for (int i = 1; i <= n; i++) read(f[i]);
  for (int i = 1; i <= m; i++) read(g[i]);
  int now = 1;
  v.push_back(0);
  for (int i = 1; i <= n; i++) {
    if (f[i] == g[now]) {
      now++;
      v.push_back(i);
    }
  }
  v.push_back(n + 1);
  if (now != m + 1) {
    cout << -1 << endl;
    return 0;
  }
  ans = 0;
  bool flag = 1;
  for (int i = 1; i < v.size(); i++) {
    if (!hhh(v[i - 1], v[i])) {
      flag = 0;
      break;
    }
  }
  if (!flag) ans = -1;
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
const int N = 2e5 + 5;
const int M = 2e3 + 5;
const int INF = 2e9;
const int sz = 15;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1);
template <class T>
inline void read(T &x) {
  char c;
  x = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') x = -1;
  T res = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
  x *= res;
}
int n, m;
long long a, b, c;
char str[N];
int f[N], g[N];
bool vis[N];
vector<int> v;
long long ans;
bool hhh(int l, int r) {
  bool flag = 1;
  for (int i = l + 1; i <= r - 1; i++) {
    if (f[i] > f[l] && f[i] > f[r]) {
      flag = 0;
      break;
    }
  }
  int remain = r - l - 1;
  if (!flag && remain < b) return false;
  ans += (1ll * remain % b) * c;
  remain = remain / b * b;
  if (a < b * c) {
    ans += (1ll * remain / b) * a;
  } else {
    if (flag) {
      ans += 1ll * remain * c;
    } else {
      ans += a;
      ans += 1ll * (remain - b) * c;
    }
  }
  return true;
}
int main() {
  read(n);
  read(m);
  read(a);
  read(b);
  read(c);
  for (int i = 1; i <= n; i++) read(f[i]);
  for (int i = 1; i <= m; i++) read(g[i]);
  int now = 1;
  v.push_back(0);
  for (int i = 1; i <= n; i++) {
    if (f[i] == g[now]) {
      now++;
      v.push_back(i);
    }
  }
  v.push_back(n + 1);
  if (now != m + 1) {
    cout << -1 << endl;
    return 0;
  }
  ans = 0;
  bool flag = 1;
  for (int i = 1; i < v.size(); i++) {
    if (!hhh(v[i - 1], v[i])) {
      flag = 0;
      break;
    }
  }
  if (!flag) ans = -1;
  cout << ans << endl;
  return 0;
}

```