#include <bits/stdc++.h>
using namespace std;
long double TL = 1.0;
bool is_tl = 0;
long long CALLS = 0;
inline bool IS() {
  if (++CALLS == 1000) {
    CALLS = 0;
    is_tl |= clock() > (TL - 0.1) * CLOCKS_PER_SEC;
  }
  return is_tl;
}
template <typename T1, typename T2>
inline void amin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void amax(T1 &a, T2 b) {
  if (a < b) a = b;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
  os << p.first << ' ' << p.second << '\n';
  return os;
}
template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &v) {
  for (auto &u : v) os << u << ' ';
  os << '\n';
  return os;
}
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 1e6 + 7, NS = 2e3 + 7, lg = 19, sq = 550, inf = 2e9 + 7,
                SM = 1e2 + 7;
const long double eps = 1e-9;
long long hash_mod[4] = {1000000007, 998244353, 1000000009, 999999937},
          mod = hash_mod[rnd() % 4];
long long hash_pows[4] = {179, 239, 1007, 2003}, P = hash_pows[rnd() % 4];
void solve() {
  long long n, m, x, y, k, ans = 0;
  cin >> n >> m >> x >> k >> y;
  vector<long long> a(n), b(m);
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long j = 0; j < m; j++) cin >> b[j];
  n += 2;
  m += 2;
  a.push_back(0);
  b.push_back(0);
  a.insert(a.begin(), -1);
  b.insert(b.begin(), -1);
  map<long long, long long> pos;
  for (long long i = 0; i < n; i++) {
    pos[a[i]] = i;
  }
  for (long long i = 0; i < m - 1; i++) {
    if (!pos.count(b[i])) {
      cout << -1;
      return;
    }
    long long cur = pos[b[i]], nxt = pos[b[i + 1]];
    if (nxt <= cur) {
      cout << -1;
      return;
    }
    long long mx = 0, len = 0;
    for (long long j = cur + 1; j < nxt; j++) {
      len++;
      amax(mx, a[j]);
    }
    bool can = 0;
    if (b[i] > mx || b[i + 1] > mx) can = 1;
    if (len < k) {
      if (!can) {
        cout << -1;
        return;
      }
      ans += len * y;
      continue;
    } else {
      ans += (len % k) * y;
      len -= len % k;
      long long mn = (len / k) * x;
      if (can) amin(mn, len * y);
      amin(mn, (len - k) * y + x);
      ans += mn;
    }
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

### Prompt

Generate 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;
long double TL = 1.0;
bool is_tl = 0;
long long CALLS = 0;
inline bool IS() {
  if (++CALLS == 1000) {
    CALLS = 0;
    is_tl |= clock() > (TL - 0.1) * CLOCKS_PER_SEC;
  }
  return is_tl;
}
template <typename T1, typename T2>
inline void amin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void amax(T1 &a, T2 b) {
  if (a < b) a = b;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
  os << p.first << ' ' << p.second << '\n';
  return os;
}
template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &v) {
  for (auto &u : v) os << u << ' ';
  os << '\n';
  return os;
}
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 1e6 + 7, NS = 2e3 + 7, lg = 19, sq = 550, inf = 2e9 + 7,
                SM = 1e2 + 7;
const long double eps = 1e-9;
long long hash_mod[4] = {1000000007, 998244353, 1000000009, 999999937},
          mod = hash_mod[rnd() % 4];
long long hash_pows[4] = {179, 239, 1007, 2003}, P = hash_pows[rnd() % 4];
void solve() {
  long long n, m, x, y, k, ans = 0;
  cin >> n >> m >> x >> k >> y;
  vector<long long> a(n), b(m);
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long j = 0; j < m; j++) cin >> b[j];
  n += 2;
  m += 2;
  a.push_back(0);
  b.push_back(0);
  a.insert(a.begin(), -1);
  b.insert(b.begin(), -1);
  map<long long, long long> pos;
  for (long long i = 0; i < n; i++) {
    pos[a[i]] = i;
  }
  for (long long i = 0; i < m - 1; i++) {
    if (!pos.count(b[i])) {
      cout << -1;
      return;
    }
    long long cur = pos[b[i]], nxt = pos[b[i + 1]];
    if (nxt <= cur) {
      cout << -1;
      return;
    }
    long long mx = 0, len = 0;
    for (long long j = cur + 1; j < nxt; j++) {
      len++;
      amax(mx, a[j]);
    }
    bool can = 0;
    if (b[i] > mx || b[i + 1] > mx) can = 1;
    if (len < k) {
      if (!can) {
        cout << -1;
        return;
      }
      ans += len * y;
      continue;
    } else {
      ans += (len % k) * y;
      len -= len % k;
      long long mn = (len / k) * x;
      if (can) amin(mn, len * y);
      amin(mn, (len - k) * y + x);
      ans += mn;
    }
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

```