#include <bits/stdc++.h>
using namespace std;
long long getcost(long long diff, long long lb, int x, int k, int y) {
  if (lb && diff >= k) {
    long long val = diff / k * x + diff % k * y;
    val = min(val, diff * y);
    return val;
  } else if (lb) {
    return diff * y;
  } else if (diff >= k) {
    long long val = diff / k * x + diff % k * y;
    if (k * y < x) {
      val = x + (diff - k) * y;
    }
    return val;
  } else {
    return -1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  int x, k, y;
  cin >> x >> k >> y;
  vector<int> a(n), b(m), bi(m);
  vector<bool> bm(m);
  for (int i = 0; i < n; i++) cin >> a[i];
  ;
  for (int i = 0; i < m; i++) {
    cin >> b[i];
  }
  int ai = 0;
  bool poss = true;
  for (int i = 0; i < m; i++) {
    int mxi = a[ai];
    while (ai < n && a[ai] != b[i]) {
      ai++;
      mxi = max(mxi, a[ai]);
    }
    if (ai == n) {
      poss = false;
      break;
    }
    bi[i] = ai;
    if (mxi == b[i] || i > 0 && mxi == b[i - 1]) {
      bm[i] = true;
    } else {
      bm[i] = false;
    }
  }
  if (!poss) {
    cout << -1;
    return 0;
  }
  long long cost = getcost(bi[0], bm[0], x, k, y);
  if (cost == -1) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i < m; i++) {
    int diff = bi[i] - bi[i - 1] - 1;
    if (diff == 0) {
      continue;
    }
    long long val = getcost(diff, bm[i], x, k, y);
    if (val == -1) {
      cout << -1;
      return 0;
    }
    cost += val;
  }
  bool lb = true;
  int diff = n - bi[m - 1] - 1;
  if (diff > 0) {
    int mxb = a[bi[m - 1]];
    for (int i = bi[m - 1] + 1; i < n; i++) {
      mxb = max(mxb, a[i]);
    }
    if (mxb != a[bi[m - 1]]) {
      lb = false;
    }
    long long val = getcost(diff, lb, x, k, y);
    if (val == -1) {
      cout << -1;
      return 0;
    }
    cost += val;
  }
  cout << cost;
  return 0;
}

### Prompt

Your task is to create 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 long getcost(long long diff, long long lb, int x, int k, int y) {
  if (lb && diff >= k) {
    long long val = diff / k * x + diff % k * y;
    val = min(val, diff * y);
    return val;
  } else if (lb) {
    return diff * y;
  } else if (diff >= k) {
    long long val = diff / k * x + diff % k * y;
    if (k * y < x) {
      val = x + (diff - k) * y;
    }
    return val;
  } else {
    return -1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  int x, k, y;
  cin >> x >> k >> y;
  vector<int> a(n), b(m), bi(m);
  vector<bool> bm(m);
  for (int i = 0; i < n; i++) cin >> a[i];
  ;
  for (int i = 0; i < m; i++) {
    cin >> b[i];
  }
  int ai = 0;
  bool poss = true;
  for (int i = 0; i < m; i++) {
    int mxi = a[ai];
    while (ai < n && a[ai] != b[i]) {
      ai++;
      mxi = max(mxi, a[ai]);
    }
    if (ai == n) {
      poss = false;
      break;
    }
    bi[i] = ai;
    if (mxi == b[i] || i > 0 && mxi == b[i - 1]) {
      bm[i] = true;
    } else {
      bm[i] = false;
    }
  }
  if (!poss) {
    cout << -1;
    return 0;
  }
  long long cost = getcost(bi[0], bm[0], x, k, y);
  if (cost == -1) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i < m; i++) {
    int diff = bi[i] - bi[i - 1] - 1;
    if (diff == 0) {
      continue;
    }
    long long val = getcost(diff, bm[i], x, k, y);
    if (val == -1) {
      cout << -1;
      return 0;
    }
    cost += val;
  }
  bool lb = true;
  int diff = n - bi[m - 1] - 1;
  if (diff > 0) {
    int mxb = a[bi[m - 1]];
    for (int i = bi[m - 1] + 1; i < n; i++) {
      mxb = max(mxb, a[i]);
    }
    if (mxb != a[bi[m - 1]]) {
      lb = false;
    }
    long long val = getcost(diff, lb, x, k, y);
    if (val == -1) {
      cout << -1;
      return 0;
    }
    cost += val;
  }
  cout << cost;
  return 0;
}

```