#include <bits/stdc++.h>
using namespace std;
long long n, m;
long long cx, cy, k;
long long a[200005];
long long b[200005];
long long app[200005];
long long kp[200005];
vector<long long> ele;
int noans = 0;
long long ans = 0;
void wk(long long x, long long y, long long big) {
  long long i;
  ele.clear();
  for (i = x; i <= y; i++) ele.push_back(a[i]);
  sort(ele.begin(), ele.end());
  if (cy * k > cx) {
    bool ok = false;
    while (ele.size() >= k) {
      ans += cx, ok = true;
      for (i = 1; i <= k; i++) ele.pop_back();
    }
    ans += ele.size() * cy;
    long long f = 0;
    if (x > 1) {
      for (i = x; i <= y; i++) {
        if (a[i] > a[x - 1]) break;
        f++;
      }
    }
    if (y < n) {
      for (i = y; i >= x; i--) {
        if (a[i] > a[y + 1]) break;
        f++;
      }
    }
    if (f < ele.size() && !ok) noans = 1;
  } else {
    long long f = 0;
    if (x > 1) {
      for (i = x; i <= y; i++) {
        if (a[i] > a[x - 1]) break;
        f++;
      }
    }
    if (y < n) {
      for (i = y; i >= x; i--) {
        if (a[i] > a[y + 1]) break;
        f++;
      }
    }
    if (f >= y - x + 1) {
      ans += ele.size() * cy;
      return;
    }
    long long tot = ele.size();
    tot -= f;
    if (tot > 0) {
      if (ele.size() >= k)
        ans += (ele.size() - k) * cy + cx;
      else
        noans = 1;
    } else
      ans += ele.size() * cy;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long i, j;
  cin >> n >> m;
  if (m > n) noans = 1;
  cin >> cx >> k >> cy;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= m; i++) cin >> b[i], app[b[i]] = i;
  long long last = 0;
  for (i = 1; i <= n; i++) {
    if (app[a[i]]) kp[i] = true;
    if (app[a[i]] && app[a[i]] < last) noans = 1;
    last = max(last, app[a[i]]);
  }
  for (i = 1; i <= n; i++)
    if (kp[i]) {
      if (i != 1) wk(1, i - 1, a[i]);
      break;
    }
  for (i = 1; i <= n; i++)
    if (kp[i]) {
      for (j = i + 1; j <= n; j++)
        if (kp[j]) break;
      if (kp[j] && i + 1 < j)
        wk(i + 1, j - 1, max(a[i], a[j]));
      else if (!kp[j] && i < n)
        wk(i + 1, n, a[i]);
    }
  if (noans)
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}

### 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;
long long n, m;
long long cx, cy, k;
long long a[200005];
long long b[200005];
long long app[200005];
long long kp[200005];
vector<long long> ele;
int noans = 0;
long long ans = 0;
void wk(long long x, long long y, long long big) {
  long long i;
  ele.clear();
  for (i = x; i <= y; i++) ele.push_back(a[i]);
  sort(ele.begin(), ele.end());
  if (cy * k > cx) {
    bool ok = false;
    while (ele.size() >= k) {
      ans += cx, ok = true;
      for (i = 1; i <= k; i++) ele.pop_back();
    }
    ans += ele.size() * cy;
    long long f = 0;
    if (x > 1) {
      for (i = x; i <= y; i++) {
        if (a[i] > a[x - 1]) break;
        f++;
      }
    }
    if (y < n) {
      for (i = y; i >= x; i--) {
        if (a[i] > a[y + 1]) break;
        f++;
      }
    }
    if (f < ele.size() && !ok) noans = 1;
  } else {
    long long f = 0;
    if (x > 1) {
      for (i = x; i <= y; i++) {
        if (a[i] > a[x - 1]) break;
        f++;
      }
    }
    if (y < n) {
      for (i = y; i >= x; i--) {
        if (a[i] > a[y + 1]) break;
        f++;
      }
    }
    if (f >= y - x + 1) {
      ans += ele.size() * cy;
      return;
    }
    long long tot = ele.size();
    tot -= f;
    if (tot > 0) {
      if (ele.size() >= k)
        ans += (ele.size() - k) * cy + cx;
      else
        noans = 1;
    } else
      ans += ele.size() * cy;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long i, j;
  cin >> n >> m;
  if (m > n) noans = 1;
  cin >> cx >> k >> cy;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= m; i++) cin >> b[i], app[b[i]] = i;
  long long last = 0;
  for (i = 1; i <= n; i++) {
    if (app[a[i]]) kp[i] = true;
    if (app[a[i]] && app[a[i]] < last) noans = 1;
    last = max(last, app[a[i]]);
  }
  for (i = 1; i <= n; i++)
    if (kp[i]) {
      if (i != 1) wk(1, i - 1, a[i]);
      break;
    }
  for (i = 1; i <= n; i++)
    if (kp[i]) {
      for (j = i + 1; j <= n; j++)
        if (kp[j]) break;
      if (kp[j] && i + 1 < j)
        wk(i + 1, j - 1, max(a[i], a[j]));
      else if (!kp[j] && i < n)
        wk(i + 1, n, a[i]);
    }
  if (noans)
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}

```