#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
const int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1},
          dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U>
inline bool chmax(T &a, U b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
inline bool chmin(T &a, U b) {
  return a > b ? (a = b, true) : false;
}
struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} iosetup;
int main() {
  int n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  vector<int> a(n), b(m);
  for (int i = (0); i < (n); ++i) cin >> a[i];
  a.emplace_back(INF);
  for (int i = (0); i < (m); ++i) cin >> b[i];
  b.emplace_back(INF);
  ++n;
  ++m;
  ll ans = 0;
  int l = -1;
  for (int i = (0); i < (m); ++i) {
    int r = l + 1;
    while (r < n && a[r] != b[i]) ++r;
    if (r == n) {
      cout << "-1\n";
      return 0;
    }
    int l_a = l == -1 ? INF : a[l];
    if (l + 1 < r) {
      bool is_small = false;
      if (l_a <= n) {
        is_small = true;
        for (int j = (l + 1); j < (r); ++j) is_small &= a[j] < l_a;
      }
      if (!is_small && a[r] <= n) {
        is_small = true;
        for (int j = (l + 1); j < (r); ++j) is_small &= a[j] < a[r];
      }
      int num = r - l - 1;
      if (num < k && !is_small) {
        cout << "-1\n";
        return 0;
      }
      if (!is_small) {
        num -= k;
        ans += x;
      }
      int b = num / k;
      ans += min(1LL * x, 1LL * k * y) * b;
      num -= b * k;
      ans += 1LL * num * y;
    }
    l = r;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
using ll = long long;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
const int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1},
          dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U>
inline bool chmax(T &a, U b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
inline bool chmin(T &a, U b) {
  return a > b ? (a = b, true) : false;
}
struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} iosetup;
int main() {
  int n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  vector<int> a(n), b(m);
  for (int i = (0); i < (n); ++i) cin >> a[i];
  a.emplace_back(INF);
  for (int i = (0); i < (m); ++i) cin >> b[i];
  b.emplace_back(INF);
  ++n;
  ++m;
  ll ans = 0;
  int l = -1;
  for (int i = (0); i < (m); ++i) {
    int r = l + 1;
    while (r < n && a[r] != b[i]) ++r;
    if (r == n) {
      cout << "-1\n";
      return 0;
    }
    int l_a = l == -1 ? INF : a[l];
    if (l + 1 < r) {
      bool is_small = false;
      if (l_a <= n) {
        is_small = true;
        for (int j = (l + 1); j < (r); ++j) is_small &= a[j] < l_a;
      }
      if (!is_small && a[r] <= n) {
        is_small = true;
        for (int j = (l + 1); j < (r); ++j) is_small &= a[j] < a[r];
      }
      int num = r - l - 1;
      if (num < k && !is_small) {
        cout << "-1\n";
        return 0;
      }
      if (!is_small) {
        num -= k;
        ans += x;
      }
      int b = num / k;
      ans += min(1LL * x, 1LL * k * y) * b;
      num -= b * k;
      ans += 1LL * num * y;
    }
    l = r;
  }
  cout << ans << '\n';
  return 0;
}

```