#include <bits/stdc++.h>
using namespace std;
int seg[(int)1e6];
int ar[(int)2e5 + 10];
int br[(int)2e5 + 10];
void build(int st, int en, int root) {
  if (st > en) return;
  if (st == en) {
    seg[root] = ar[st];
    return;
  }
  int mid = (st + en) / 2;
  build(st, mid, 2 * root + 1);
  build(mid + 1, en, 2 * root + 2);
  seg[root] = max(seg[2 * root + 1], seg[2 * root + 2]);
}
int query(int st, int en, int qs, int qe, int root) {
  if (st > qe || en < qs || st > en) {
    return 0;
  }
  if (st >= qs and en <= qe) {
    return seg[root];
  }
  int mid = (st + en) / 2;
  return max(query(st, mid, qs, qe, 2 * root + 1),
             query(mid + 1, en, qs, qe, 2 * root + 2));
}
int main() {
  int64_t n, m;
  cin >> n >> m;
  int64_t x, k, y;
  cin >> x >> k >> y;
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
  }
  for (int i = 1; i <= m; i++) {
    cin >> br[i];
  }
  br[0] = 0;
  br[m + 1] = -1;
  ar[n] = -1;
  build(0, n - 1, 0);
  int64_t pos = 1;
  int64_t ini = 0;
  int64_t ans = 0;
  bool flag = false;
  for (int i = 0; i <= n; i++) {
    if (ar[i] == br[pos]) {
      int64_t len = i - ini;
      if (len > 0) {
        if (len >= k) {
          if (y * k < x) {
            int qr = query(0, n - 1, ini, i - 1, 0);
            if (qr < br[pos] || qr < br[pos - 1]) {
              ans += len * y;
            } else {
              ans += x;
              ans += (len - k) * y;
            }
          } else {
            ans += y * (len % k);
            ans += x * (len / k);
          }
        } else {
          int qr = query(0, n - 1, ini, i - 1, 0);
          if (br[pos] < qr and br[pos - 1] < qr) {
            flag = false;
            break;
          } else {
            ans += len * y;
          }
        }
      }
      ini = i + 1;
      pos++;
      if (pos == m + 2) {
        flag = true;
        break;
      }
    }
  }
  if (flag) {
    cout << ans << "\n";
  } else {
    cout << "-1\n";
  }
  return 0;
}

### 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;
int seg[(int)1e6];
int ar[(int)2e5 + 10];
int br[(int)2e5 + 10];
void build(int st, int en, int root) {
  if (st > en) return;
  if (st == en) {
    seg[root] = ar[st];
    return;
  }
  int mid = (st + en) / 2;
  build(st, mid, 2 * root + 1);
  build(mid + 1, en, 2 * root + 2);
  seg[root] = max(seg[2 * root + 1], seg[2 * root + 2]);
}
int query(int st, int en, int qs, int qe, int root) {
  if (st > qe || en < qs || st > en) {
    return 0;
  }
  if (st >= qs and en <= qe) {
    return seg[root];
  }
  int mid = (st + en) / 2;
  return max(query(st, mid, qs, qe, 2 * root + 1),
             query(mid + 1, en, qs, qe, 2 * root + 2));
}
int main() {
  int64_t n, m;
  cin >> n >> m;
  int64_t x, k, y;
  cin >> x >> k >> y;
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
  }
  for (int i = 1; i <= m; i++) {
    cin >> br[i];
  }
  br[0] = 0;
  br[m + 1] = -1;
  ar[n] = -1;
  build(0, n - 1, 0);
  int64_t pos = 1;
  int64_t ini = 0;
  int64_t ans = 0;
  bool flag = false;
  for (int i = 0; i <= n; i++) {
    if (ar[i] == br[pos]) {
      int64_t len = i - ini;
      if (len > 0) {
        if (len >= k) {
          if (y * k < x) {
            int qr = query(0, n - 1, ini, i - 1, 0);
            if (qr < br[pos] || qr < br[pos - 1]) {
              ans += len * y;
            } else {
              ans += x;
              ans += (len - k) * y;
            }
          } else {
            ans += y * (len % k);
            ans += x * (len / k);
          }
        } else {
          int qr = query(0, n - 1, ini, i - 1, 0);
          if (br[pos] < qr and br[pos - 1] < qr) {
            flag = false;
            break;
          } else {
            ans += len * y;
          }
        }
      }
      ini = i + 1;
      pos++;
      if (pos == m + 2) {
        flag = true;
        break;
      }
    }
  }
  if (flag) {
    cout << ans << "\n";
  } else {
    cout << "-1\n";
  }
  return 0;
}

```