#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
const long long INF = 1e17 + 5;
int a[MAXN];
int b[MAXN];
long long x, k, y;
bool check_valid(int n, int m) {
  int ptr = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[ptr]) ptr++;
    if (ptr == m) break;
  }
  return ptr >= m;
}
long long min_clear(int sz, int mx, int nxt) {
  if (sz == 0) return 0;
  if (mx > nxt && sz < k) return -1;
  vector<long long> dp(sz + 5, 0);
  dp[1] = INF;
  if (mx < nxt) dp[1] = min(dp[1], y);
  if (k == 1) dp[1] = min(dp[1], x);
  for (int s = 2; s <= sz; s++) {
    dp[s] = INF;
    if (s >= k) dp[s] = min(dp[s], x + dp[s - k]);
    dp[s] = min(dp[s], y + dp[s - 1]);
  }
  return dp[sz];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m >> x >> k >> y;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];
  if (!check_valid(n, m)) {
    cout << -1 << "\n";
    return 0;
  }
  vector<int> pending;
  int ptr = 0;
  bool ok = true;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] != b[ptr]) {
      pending.push_back(a[i]);
      continue;
    }
    int sz = pending.size();
    int mx =
        (pending.empty() ? 0 : *max_element(pending.begin(), pending.end()));
    int prv = (ptr == 0 ? 0 : b[ptr - 1]);
    int nxt = max(prv, a[i]);
    long long res = min_clear(sz, mx, nxt);
    if (res == -1) {
      ok = false;
      break;
    } else {
      ans += res;
    }
    pending.clear();
    ptr++;
  }
  if (!pending.empty()) {
    int sz = pending.size();
    int mx =
        (pending.empty() ? 0 : *max_element(pending.begin(), pending.end()));
    int nxt = (ptr == 0 ? 0 : b[ptr - 1]);
    long long res = min_clear(sz, mx, nxt);
    if (res == -1) {
      ok = false;
    } else
      ans += res;
  }
  cout << (ok ? ans : -1) << "\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;
const int MAXN = 200005;
const long long INF = 1e17 + 5;
int a[MAXN];
int b[MAXN];
long long x, k, y;
bool check_valid(int n, int m) {
  int ptr = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[ptr]) ptr++;
    if (ptr == m) break;
  }
  return ptr >= m;
}
long long min_clear(int sz, int mx, int nxt) {
  if (sz == 0) return 0;
  if (mx > nxt && sz < k) return -1;
  vector<long long> dp(sz + 5, 0);
  dp[1] = INF;
  if (mx < nxt) dp[1] = min(dp[1], y);
  if (k == 1) dp[1] = min(dp[1], x);
  for (int s = 2; s <= sz; s++) {
    dp[s] = INF;
    if (s >= k) dp[s] = min(dp[s], x + dp[s - k]);
    dp[s] = min(dp[s], y + dp[s - 1]);
  }
  return dp[sz];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m >> x >> k >> y;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];
  if (!check_valid(n, m)) {
    cout << -1 << "\n";
    return 0;
  }
  vector<int> pending;
  int ptr = 0;
  bool ok = true;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] != b[ptr]) {
      pending.push_back(a[i]);
      continue;
    }
    int sz = pending.size();
    int mx =
        (pending.empty() ? 0 : *max_element(pending.begin(), pending.end()));
    int prv = (ptr == 0 ? 0 : b[ptr - 1]);
    int nxt = max(prv, a[i]);
    long long res = min_clear(sz, mx, nxt);
    if (res == -1) {
      ok = false;
      break;
    } else {
      ans += res;
    }
    pending.clear();
    ptr++;
  }
  if (!pending.empty()) {
    int sz = pending.size();
    int mx =
        (pending.empty() ? 0 : *max_element(pending.begin(), pending.end()));
    int nxt = (ptr == 0 ? 0 : b[ptr - 1]);
    long long res = min_clear(sz, mx, nxt);
    if (res == -1) {
      ok = false;
    } else
      ans += res;
  }
  cout << (ok ? ans : -1) << "\n";
  return 0;
}

```