#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
long long n, m, x, k, y;
vector<long long> a, b, pos;
long long destroy(long long lx, long long rx) {
  if (lx > rx) return 0;
  long long torem = rx - lx + 1;
  long long maxi = 0, maxpos, lhs = max((long long)0, lx - 1),
            rhs = min(rx + 1, n - 1);
  for (long long i = lhs; i <= rhs; i++) {
    if (a[i] > maxi) {
      maxi = a[i];
      maxpos = i;
    }
  }
  long long ans = pow(10, 18);
  long long p = torem / k;
  bool con = (maxpos >= lx && maxpos <= rx);
  long long start;
  if (con)
    start = 1;
  else
    start = 0;
  for (long long i = start; i <= p; i++) {
    long long cost = i * x + y * (torem - i * k);
    ans = min(ans, cost);
  }
  if (con && torem < k) return -1;
  return ans;
}
void aksayushx() {
  cin >> n >> m;
  cin >> x >> k >> y;
  a.resize(n);
  b.resize(m);
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < m; i++) cin >> b[i];
  long long look = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] == b[look]) {
      pos.push_back(i);
      look++;
    }
    if (look >= m) break;
  }
  long long ans = 0, len = pos.size() - 1;
  if (pos.size() < m) {
    cout << -1;
    return;
  }
  if (pos[0] != 0) ans += destroy(0, pos[0] - 1);
  bool ok = true;
  if (ans == -1) ok = false;
  for (long long i = 0; i < len; i++) {
    long long cost = destroy(pos[i] + 1, pos[i + 1] - 1);
    ans += cost;
    if (cost == -1) {
      ok = false;
      break;
    }
  }
  if (pos[len] != n - 1) {
    long long cost = destroy(pos[len] + 1, n - 1);
    ans += cost;
    if (cost == -1) ok = false;
  }
  if (!ok)
    cout << -1;
  else
    cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) aksayushx();
  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>
#pragma GCC optimize("O3")
using namespace std;
long long n, m, x, k, y;
vector<long long> a, b, pos;
long long destroy(long long lx, long long rx) {
  if (lx > rx) return 0;
  long long torem = rx - lx + 1;
  long long maxi = 0, maxpos, lhs = max((long long)0, lx - 1),
            rhs = min(rx + 1, n - 1);
  for (long long i = lhs; i <= rhs; i++) {
    if (a[i] > maxi) {
      maxi = a[i];
      maxpos = i;
    }
  }
  long long ans = pow(10, 18);
  long long p = torem / k;
  bool con = (maxpos >= lx && maxpos <= rx);
  long long start;
  if (con)
    start = 1;
  else
    start = 0;
  for (long long i = start; i <= p; i++) {
    long long cost = i * x + y * (torem - i * k);
    ans = min(ans, cost);
  }
  if (con && torem < k) return -1;
  return ans;
}
void aksayushx() {
  cin >> n >> m;
  cin >> x >> k >> y;
  a.resize(n);
  b.resize(m);
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < m; i++) cin >> b[i];
  long long look = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] == b[look]) {
      pos.push_back(i);
      look++;
    }
    if (look >= m) break;
  }
  long long ans = 0, len = pos.size() - 1;
  if (pos.size() < m) {
    cout << -1;
    return;
  }
  if (pos[0] != 0) ans += destroy(0, pos[0] - 1);
  bool ok = true;
  if (ans == -1) ok = false;
  for (long long i = 0; i < len; i++) {
    long long cost = destroy(pos[i] + 1, pos[i + 1] - 1);
    ans += cost;
    if (cost == -1) {
      ok = false;
      break;
    }
  }
  if (pos[len] != n - 1) {
    long long cost = destroy(pos[len] + 1, n - 1);
    ans += cost;
    if (cost == -1) ok = false;
  }
  if (!ok)
    cout << -1;
  else
    cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) aksayushx();
  return 0;
}

```