#include <bits/stdc++.h>
using namespace std;
int main() {
  unsigned long long n, m;
  cin >> n >> m;
  unsigned long long x, k, y;
  cin >> x >> k >> y;
  vector<unsigned long long> arr, brr;
  map<unsigned long long, unsigned long long> umap;
  for (unsigned long long i = 0; i < n; i++) {
    unsigned long long a;
    cin >> a;
    arr.push_back(a);
    umap[a] = i;
  }
  for (unsigned long long i = 1; i <= m; i++) {
    unsigned long long a;
    cin >> a;
    brr.push_back(a);
  }
  if (arr.size() == brr.size() && arr == brr) {
    cout << 0 << endl;
    return 0;
  }
  if (arr.size() == brr.size() && arr != brr) {
    cout << -1 << endl;
    return 0;
  }
  unsigned long long sum = 0, st = 0, tc = 0, mc = 0, et = 0;
  for (unsigned long long i = 0; i < n; i++) {
    if (et == brr.size()) {
      if (arr[i] >= brr[et - 1]) {
        mc++;
        tc++;
      } else {
        tc++;
      }
      continue;
    }
    if (arr[i] == brr[et]) {
      if (tc >= k) {
        if (mc == 0)
          sum = sum + min(x * (tc / k) + y * (tc % k), y * (tc));
        else
          sum = sum + min(x * (tc / k) + y * (tc % k), y * (tc - k) + x);
      } else {
        if (mc != 0) {
          cout << -1 << endl;
          return 0;
        } else {
          sum = sum + y * tc;
        }
      }
      st = et;
      et++;
      tc = 0;
      mc = 0;
      continue;
    }
    if (arr[i] >= brr[st] && arr[i] >= brr[et]) {
      mc++;
      tc++;
    } else
      tc++;
  }
  if (et != brr.size()) {
    cout << -1 << endl;
    return 0;
  }
  if (tc >= k) {
    if (mc == 0)
      sum = sum + min(x * (tc / k) + y * (tc % k), y * (tc));
    else
      sum = sum + min(x * (tc / k) + y * (tc % k), y * (tc - k) + x);
  } else {
    if (mc != 0) {
      cout << -1 << endl;
      return 0;
    } else {
      sum = sum + y * tc;
    }
  }
  cout << sum << endl;
}

### Prompt

Please formulate 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;
int main() {
  unsigned long long n, m;
  cin >> n >> m;
  unsigned long long x, k, y;
  cin >> x >> k >> y;
  vector<unsigned long long> arr, brr;
  map<unsigned long long, unsigned long long> umap;
  for (unsigned long long i = 0; i < n; i++) {
    unsigned long long a;
    cin >> a;
    arr.push_back(a);
    umap[a] = i;
  }
  for (unsigned long long i = 1; i <= m; i++) {
    unsigned long long a;
    cin >> a;
    brr.push_back(a);
  }
  if (arr.size() == brr.size() && arr == brr) {
    cout << 0 << endl;
    return 0;
  }
  if (arr.size() == brr.size() && arr != brr) {
    cout << -1 << endl;
    return 0;
  }
  unsigned long long sum = 0, st = 0, tc = 0, mc = 0, et = 0;
  for (unsigned long long i = 0; i < n; i++) {
    if (et == brr.size()) {
      if (arr[i] >= brr[et - 1]) {
        mc++;
        tc++;
      } else {
        tc++;
      }
      continue;
    }
    if (arr[i] == brr[et]) {
      if (tc >= k) {
        if (mc == 0)
          sum = sum + min(x * (tc / k) + y * (tc % k), y * (tc));
        else
          sum = sum + min(x * (tc / k) + y * (tc % k), y * (tc - k) + x);
      } else {
        if (mc != 0) {
          cout << -1 << endl;
          return 0;
        } else {
          sum = sum + y * tc;
        }
      }
      st = et;
      et++;
      tc = 0;
      mc = 0;
      continue;
    }
    if (arr[i] >= brr[st] && arr[i] >= brr[et]) {
      mc++;
      tc++;
    } else
      tc++;
  }
  if (et != brr.size()) {
    cout << -1 << endl;
    return 0;
  }
  if (tc >= k) {
    if (mc == 0)
      sum = sum + min(x * (tc / k) + y * (tc % k), y * (tc));
    else
      sum = sum + min(x * (tc / k) + y * (tc % k), y * (tc - k) + x);
  } else {
    if (mc != 0) {
      cout << -1 << endl;
      return 0;
    } else {
      sum = sum + y * tc;
    }
  }
  cout << sum << endl;
}

```