#include <bits/stdc++.h>
using namespace std;
const long long INF = LLONG_MAX;
const int mxn = 2e5 + 5;
bool DEBUG = 0;
int n, m, X, Y, k;
int a[mxn], b[mxn];
pair<int, int> clr[mxn];
unordered_map<int, int> pos;
int main() {
  cin >> n >> m >> X >> k >> Y;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  bool ok = 1;
  for (int i = 0; i < m; i++) {
    cin >> b[i];
  }
  clr[0] = {-1, pos[b[0]]};
  for (int i = 1; i < m; i++) {
    if (pos[b[i]] < pos[b[i - 1]])
      ok = 0;
    else
      clr[i] = {pos[b[i - 1]], pos[b[i]]};
  }
  clr[m] = {pos[b[m - 1]], n};
  if (!ok) {
    cout << -1 << endl;
    return 0;
  }
  long long ans = 0ll;
  for (int i = 0; i <= m && ok; i++) {
    int le = clr[i].first;
    int ri = clr[i].second;
    int cnt = (ri - 1) - (le + 1) + 1;
    if (cnt == 0) continue;
    if (cnt >= k) {
      if (X < Y * k) {
        int x = cnt / k;
        int y = cnt % k;
        ans += 1ll * x * X + 1ll * y * Y;
      } else {
        vector<int> bigger;
        int val = max(le == -1 ? 0 : a[le], a[ri]);
        for (int i = le + 1; i < ri; i++) {
          if (a[i] > val) bigger.emplace_back(i);
        }
        if (bigger.empty()) {
          ans += 1ll * cnt * Y;
          continue;
        }
        int x = 1;
        int y = cnt - x * k;
        ans += 1ll * x * X + 1ll * y * Y;
      }
    } else {
      vector<int> bigger;
      int val = max(le == -1 ? 0 : a[le], a[ri]);
      for (int i = le + 1; i < ri; i++) {
        if (a[i] > val) bigger.emplace_back(i);
      }
      if (!bigger.empty())
        ok = 0;
      else
        ans += 1ll * cnt * Y;
    }
  }
  if (!ok) ans = -1;
  cout << ans << endl;
}

### Prompt

Develop a solution in cpp 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>
using namespace std;
const long long INF = LLONG_MAX;
const int mxn = 2e5 + 5;
bool DEBUG = 0;
int n, m, X, Y, k;
int a[mxn], b[mxn];
pair<int, int> clr[mxn];
unordered_map<int, int> pos;
int main() {
  cin >> n >> m >> X >> k >> Y;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  bool ok = 1;
  for (int i = 0; i < m; i++) {
    cin >> b[i];
  }
  clr[0] = {-1, pos[b[0]]};
  for (int i = 1; i < m; i++) {
    if (pos[b[i]] < pos[b[i - 1]])
      ok = 0;
    else
      clr[i] = {pos[b[i - 1]], pos[b[i]]};
  }
  clr[m] = {pos[b[m - 1]], n};
  if (!ok) {
    cout << -1 << endl;
    return 0;
  }
  long long ans = 0ll;
  for (int i = 0; i <= m && ok; i++) {
    int le = clr[i].first;
    int ri = clr[i].second;
    int cnt = (ri - 1) - (le + 1) + 1;
    if (cnt == 0) continue;
    if (cnt >= k) {
      if (X < Y * k) {
        int x = cnt / k;
        int y = cnt % k;
        ans += 1ll * x * X + 1ll * y * Y;
      } else {
        vector<int> bigger;
        int val = max(le == -1 ? 0 : a[le], a[ri]);
        for (int i = le + 1; i < ri; i++) {
          if (a[i] > val) bigger.emplace_back(i);
        }
        if (bigger.empty()) {
          ans += 1ll * cnt * Y;
          continue;
        }
        int x = 1;
        int y = cnt - x * k;
        ans += 1ll * x * X + 1ll * y * Y;
      }
    } else {
      vector<int> bigger;
      int val = max(le == -1 ? 0 : a[le], a[ri]);
      for (int i = le + 1; i < ri; i++) {
        if (a[i] > val) bigger.emplace_back(i);
      }
      if (!bigger.empty())
        ok = 0;
      else
        ans += 1ll * cnt * Y;
    }
  }
  if (!ok) ans = -1;
  cout << ans << endl;
}

```