#include <bits/stdc++.h>
const int N = 200010;
const long long INF = 1ll << 60;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
int t;
int n, m;
int a[N];
int b[N];
int id[N], c[N];
int x, k, y;
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%d%d%d", &x, &k, &y);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    id[a[i]] = i;
  }
  a[0] = a[n + 1] = 0;
  bool flag = true;
  c[0] = 0;
  c[m + 1] = n + 1;
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &b[i]);
    c[i] = id[b[i]];
    if (c[i] < c[i - 1]) flag = false;
  }
  if (!flag) {
    printf("-1\n");
    return 0;
  }
  for (int i = 0; i <= m; ++i) {
    int l = c[i];
    int r = c[i + 1];
    assert(l < r);
    if (l + 1 == r) continue;
    if (r - l - 1 < k) {
      int mx = 0, id_mx = -1;
      for (int j = l; j <= r; ++j) {
        if (a[j] > mx) mx = a[j], id_mx = j;
      }
      if (id_mx != l && id_mx != r) {
        flag = false;
        break;
      } else
        ans += 1ll * y * (r - l - 1);
    } else {
      if (1ll * y * k < x) {
        int mx = 0, id_mx = -1;
        for (int j = l; j <= r; ++j) {
          if (a[j] > mx) mx = a[j], id_mx = j;
        }
        if (id_mx != l && id_mx != r) {
          ans += (1ll * x + 1ll * y * (r - l - 1 - k));
        } else {
          ans += 1ll * y * (r - l - 1);
        }
      } else {
        int tmp = (r - l - 1) / k;
        int tmp2 = (r - l - 1) % k;
        ans += (1ll * tmp * x + 1ll * tmp2 * y);
      }
    }
  }
  if (!flag)
    printf("-1\n");
  else
    printf("%lld\n", ans);
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
const int N = 200010;
const long long INF = 1ll << 60;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
int t;
int n, m;
int a[N];
int b[N];
int id[N], c[N];
int x, k, y;
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%d%d%d", &x, &k, &y);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    id[a[i]] = i;
  }
  a[0] = a[n + 1] = 0;
  bool flag = true;
  c[0] = 0;
  c[m + 1] = n + 1;
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &b[i]);
    c[i] = id[b[i]];
    if (c[i] < c[i - 1]) flag = false;
  }
  if (!flag) {
    printf("-1\n");
    return 0;
  }
  for (int i = 0; i <= m; ++i) {
    int l = c[i];
    int r = c[i + 1];
    assert(l < r);
    if (l + 1 == r) continue;
    if (r - l - 1 < k) {
      int mx = 0, id_mx = -1;
      for (int j = l; j <= r; ++j) {
        if (a[j] > mx) mx = a[j], id_mx = j;
      }
      if (id_mx != l && id_mx != r) {
        flag = false;
        break;
      } else
        ans += 1ll * y * (r - l - 1);
    } else {
      if (1ll * y * k < x) {
        int mx = 0, id_mx = -1;
        for (int j = l; j <= r; ++j) {
          if (a[j] > mx) mx = a[j], id_mx = j;
        }
        if (id_mx != l && id_mx != r) {
          ans += (1ll * x + 1ll * y * (r - l - 1 - k));
        } else {
          ans += 1ll * y * (r - l - 1);
        }
      } else {
        int tmp = (r - l - 1) / k;
        int tmp2 = (r - l - 1) % k;
        ans += (1ll * tmp * x + 1ll * tmp2 * y);
      }
    }
  }
  if (!flag)
    printf("-1\n");
  else
    printf("%lld\n", ans);
  return 0;
}

```