#include <bits/stdc++.h>
using namespace std;
const int maxn = 2 * 1e5;
const int mx = (1 << 20) + 99;
long long n, m, x, s, y, a[maxn + 10], aans, ans, flag;
struct lll {
  long long num, pos;
} b[maxn + 10];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int t;
  cin >> n >> m;
  cin >> x >> s >> y;
  for (int i = 1; i <= n; i++) cin >> a[i];
  flag = 0;
  for (int i = 1, j = 1; j <= m; j++) {
    cin >> b[j].num;
    if (i > n) flag = 1;
    if (!flag)
      for (; i <= n;) {
        if (a[i] == b[j].num) {
          b[j].pos = i;
          i++;
          break;
        }
        if (i == n) flag = 1;
        i++;
      }
  }
  if (flag) {
    cout << "-1" << endl;
    return 0;
  }
  ans = 0;
  for (int i = 1, j = 1, k; i <= m + 1; i++) {
    if (i == m + 1)
      k = n;
    else
      k = b[i].pos - 1;
    if (i != 1) j = b[i - 1].pos + 1;
    if (k < j) continue;
    aans = -1;
    for (int h = j; h <= k; h++) aans = max(aans, a[h]);
    if (k - j + 1 >= s) {
      long long ss;
      ss = min(((k - j + 1) % s) * y + ((k - j + 1) / s) * x,
               x + (k - j + 1 - s) * y);
      if (i == 1) {
        if (b[1].num > aans) ss = min((k - j + 1) * y, ss);
      } else if (i == m + 1) {
        if (b[m].num > aans) ss = min((k - j + 1) * y, ss);
      } else {
        if (b[i - 1].num > aans || b[i].num > aans)
          ss = min((k - j + 1) * y, ss);
      }
      ans += ss;
      continue;
    }
    if (i == 1) {
      if (b[1].num < aans)
        flag = 1;
      else
        ans += ((k - j + 1) % s) * y;
    } else if (i == m + 1) {
      if (b[m].num < aans)
        flag = 1;
      else
        ans += ((k - j + 1) % s) * y;
    } else {
      if (b[i - 1].num < aans && b[i].num < aans)
        flag = 1;
      else
        ans += ((k - j + 1) % s) * y;
    }
    if (flag) break;
  }
  if (flag)
    cout << "-1" << endl;
  else
    cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
const int maxn = 2 * 1e5;
const int mx = (1 << 20) + 99;
long long n, m, x, s, y, a[maxn + 10], aans, ans, flag;
struct lll {
  long long num, pos;
} b[maxn + 10];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int t;
  cin >> n >> m;
  cin >> x >> s >> y;
  for (int i = 1; i <= n; i++) cin >> a[i];
  flag = 0;
  for (int i = 1, j = 1; j <= m; j++) {
    cin >> b[j].num;
    if (i > n) flag = 1;
    if (!flag)
      for (; i <= n;) {
        if (a[i] == b[j].num) {
          b[j].pos = i;
          i++;
          break;
        }
        if (i == n) flag = 1;
        i++;
      }
  }
  if (flag) {
    cout << "-1" << endl;
    return 0;
  }
  ans = 0;
  for (int i = 1, j = 1, k; i <= m + 1; i++) {
    if (i == m + 1)
      k = n;
    else
      k = b[i].pos - 1;
    if (i != 1) j = b[i - 1].pos + 1;
    if (k < j) continue;
    aans = -1;
    for (int h = j; h <= k; h++) aans = max(aans, a[h]);
    if (k - j + 1 >= s) {
      long long ss;
      ss = min(((k - j + 1) % s) * y + ((k - j + 1) / s) * x,
               x + (k - j + 1 - s) * y);
      if (i == 1) {
        if (b[1].num > aans) ss = min((k - j + 1) * y, ss);
      } else if (i == m + 1) {
        if (b[m].num > aans) ss = min((k - j + 1) * y, ss);
      } else {
        if (b[i - 1].num > aans || b[i].num > aans)
          ss = min((k - j + 1) * y, ss);
      }
      ans += ss;
      continue;
    }
    if (i == 1) {
      if (b[1].num < aans)
        flag = 1;
      else
        ans += ((k - j + 1) % s) * y;
    } else if (i == m + 1) {
      if (b[m].num < aans)
        flag = 1;
      else
        ans += ((k - j + 1) % s) * y;
    } else {
      if (b[i - 1].num < aans && b[i].num < aans)
        flag = 1;
      else
        ans += ((k - j + 1) % s) * y;
    }
    if (flag) break;
  }
  if (flag)
    cout << "-1" << endl;
  else
    cout << ans << endl;
  return 0;
}

```