#include <bits/stdc++.h>
using namespace std;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long NINF = 0xc0c0c0c0c0c0c0c0;
inline void read(long long &x) {
  bool flag = false;
  x = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') flag = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  x = flag ? -x : x;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const long long mod = 1e9 + 7;
const long long maxn = (2e5 + 5);
long long a[maxn], b[maxn], pos[maxn];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  for (long long i = (1); i <= (n); ++i) cin >> a[i];
  for (long long i = (1); i <= (m); ++i) cin >> b[i];
  long long np = 1;
  for (long long i = (1); i <= (n); ++i) {
    if (a[i] == b[np]) pos[np++] = i;
  }
  if (np <= m) {
    cout << -1 << "\n";
    return 0;
  }
  long long ans = 0;
  pos[0] = 0;
  pos[np] = n + 1;
  for (long long i = (1); i <= (np); ++i) {
    long long beg = pos[i - 1], ed = pos[i];
    long long maxv = max(a[beg], a[ed]);
    vector<long long> big;
    for (long long j = (beg + 1); j <= (ed - 1); ++j) {
      if (a[j] > maxv) big.push_back(j);
    }
    if (big.empty()) {
      if (y * k <= x)
        ans += y * (ed - beg - 1);
      else {
        ans += (ed - beg - 1) / k * x + (ed - beg - 1) % k * y;
      }
    } else {
      if (ed - beg - 1 < k) {
        cout << -1 << "\n";
        return 0;
      } else if (y * k >= x) {
        ans += (ed - beg - 1) / k * x + (ed - beg - 1) % k * y;
      } else {
        ans += x + (ed - beg - 1 - k) * y;
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long NINF = 0xc0c0c0c0c0c0c0c0;
inline void read(long long &x) {
  bool flag = false;
  x = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') flag = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  x = flag ? -x : x;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const long long mod = 1e9 + 7;
const long long maxn = (2e5 + 5);
long long a[maxn], b[maxn], pos[maxn];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  for (long long i = (1); i <= (n); ++i) cin >> a[i];
  for (long long i = (1); i <= (m); ++i) cin >> b[i];
  long long np = 1;
  for (long long i = (1); i <= (n); ++i) {
    if (a[i] == b[np]) pos[np++] = i;
  }
  if (np <= m) {
    cout << -1 << "\n";
    return 0;
  }
  long long ans = 0;
  pos[0] = 0;
  pos[np] = n + 1;
  for (long long i = (1); i <= (np); ++i) {
    long long beg = pos[i - 1], ed = pos[i];
    long long maxv = max(a[beg], a[ed]);
    vector<long long> big;
    for (long long j = (beg + 1); j <= (ed - 1); ++j) {
      if (a[j] > maxv) big.push_back(j);
    }
    if (big.empty()) {
      if (y * k <= x)
        ans += y * (ed - beg - 1);
      else {
        ans += (ed - beg - 1) / k * x + (ed - beg - 1) % k * y;
      }
    } else {
      if (ed - beg - 1 < k) {
        cout << -1 << "\n";
        return 0;
      } else if (y * k >= x) {
        ans += (ed - beg - 1) / k * x + (ed - beg - 1) % k * y;
      } else {
        ans += x + (ed - beg - 1 - k) * y;
      }
    }
  }
  cout << ans;
  return 0;
}

```