#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using db = double;
using vi = vector<int>;
const int inf = 0x3f3f3f3f;
const db eps = 1e-8;
const int mod = 1e9 + 7;
ll qpow(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
int n, m, x, k, y;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> x >> k >> y;
  vi a(n + 2), b(m);
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) cin >> b[i];
  vi p{0};
  for (int i = 1, j = 0; i <= n; ++i) {
    if (j < m && a[i] == b[j]) {
      p.push_back(i);
      ++j;
    }
  }
  if (((int)(p).size()) < m + 1) {
    cout << "-1\n";
    return 0;
  }
  p.push_back(n + 1);
  ll ans = 0;
  for (int i = 0; i < m + 1; ++i) {
    int l = p[i], r = p[i + 1];
    int len = r - l - 1;
    if (len == 0) continue;
    int mx = *max_element(a.begin() + l + 1, a.begin() + r);
    if (len < k) {
      if (max(a[l], a[r]) < mx) {
        cout << "-1\n";
        return 0;
      }
      ans += 1ll * len * y;
    } else if (x < 1ll * k * y) {
      ans += 1ll * (len % k) * y + 1ll * (len / k) * x;
    } else if (max(a[l], a[r]) < mx) {
      ans += x + 1ll * (len - k) * y;
    } else {
      ans += 1ll * len * y;
    }
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Please formulate 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;
using ll = long long;
using pii = pair<int, int>;
using db = double;
using vi = vector<int>;
const int inf = 0x3f3f3f3f;
const db eps = 1e-8;
const int mod = 1e9 + 7;
ll qpow(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
int n, m, x, k, y;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> x >> k >> y;
  vi a(n + 2), b(m);
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) cin >> b[i];
  vi p{0};
  for (int i = 1, j = 0; i <= n; ++i) {
    if (j < m && a[i] == b[j]) {
      p.push_back(i);
      ++j;
    }
  }
  if (((int)(p).size()) < m + 1) {
    cout << "-1\n";
    return 0;
  }
  p.push_back(n + 1);
  ll ans = 0;
  for (int i = 0; i < m + 1; ++i) {
    int l = p[i], r = p[i + 1];
    int len = r - l - 1;
    if (len == 0) continue;
    int mx = *max_element(a.begin() + l + 1, a.begin() + r);
    if (len < k) {
      if (max(a[l], a[r]) < mx) {
        cout << "-1\n";
        return 0;
      }
      ans += 1ll * len * y;
    } else if (x < 1ll * k * y) {
      ans += 1ll * (len % k) * y + 1ll * (len / k) * x;
    } else if (max(a[l], a[r]) < mx) {
      ans += x + 1ll * (len - k) * y;
    } else {
      ans += 1ll * len * y;
    }
  }
  cout << ans << "\n";
  return 0;
}

```