#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
template <class T>
void read(T &val) {
  T x = 0;
  T bz = 1;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  if (c == '-') {
    bz = -1;
    c = getchar();
  }
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - 48;
  val = x * bz;
}
const int maxn = 1e6 + 11;
const int mod = 1e9 + 7;
int t;
long long x, a[maxn], n, m, b[maxn];
bool mpp[maxn];
bool vis[maxn];
int p[maxn];
int main() {
  read(n);
  read(m);
  long long x, y, k;
  read(x);
  read(k);
  read(y);
  for (int i = 1; i <= n; i++) read(a[i]), p[a[i]] = i;
  for (int i = 1; i <= m; i++) read(b[i]), mpp[b[i]] = true;
  for (int i = 1; i <= n; i++) {
    if (!mpp[a[i]]) vis[i] = true;
  }
  int pos1 = 1;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    if (a[i] != b[pos1]) {
      puts("-1");
      return 0;
    }
    pos1++;
  }
  long long ans = 0;
  int xx = 1;
  while (xx <= n) {
    if (vis[xx] == false) {
      xx++;
      continue;
    }
    long long cnt = 0, mx = 0;
    int l = xx - 1;
    while (vis[xx]) {
      cnt++;
      xx++;
      mx = max(mx, a[xx - 1]);
    }
    if (cnt >= k) {
      if (a[l] > mx || a[xx] > mx) {
        ans += cnt / k * min(x, k * y);
      } else {
        cnt -= k;
        ans += x;
        ans += cnt / k * min(x, k * y);
      }
      ans += cnt % k * y;
    } else {
      if (a[l] < mx && mx > a[xx]) {
        puts("-1");
        return 0;
      }
      ans += cnt * y;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please formulate 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 INF = 0x3f3f3f3f;
template <class T>
void read(T &val) {
  T x = 0;
  T bz = 1;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  if (c == '-') {
    bz = -1;
    c = getchar();
  }
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - 48;
  val = x * bz;
}
const int maxn = 1e6 + 11;
const int mod = 1e9 + 7;
int t;
long long x, a[maxn], n, m, b[maxn];
bool mpp[maxn];
bool vis[maxn];
int p[maxn];
int main() {
  read(n);
  read(m);
  long long x, y, k;
  read(x);
  read(k);
  read(y);
  for (int i = 1; i <= n; i++) read(a[i]), p[a[i]] = i;
  for (int i = 1; i <= m; i++) read(b[i]), mpp[b[i]] = true;
  for (int i = 1; i <= n; i++) {
    if (!mpp[a[i]]) vis[i] = true;
  }
  int pos1 = 1;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    if (a[i] != b[pos1]) {
      puts("-1");
      return 0;
    }
    pos1++;
  }
  long long ans = 0;
  int xx = 1;
  while (xx <= n) {
    if (vis[xx] == false) {
      xx++;
      continue;
    }
    long long cnt = 0, mx = 0;
    int l = xx - 1;
    while (vis[xx]) {
      cnt++;
      xx++;
      mx = max(mx, a[xx - 1]);
    }
    if (cnt >= k) {
      if (a[l] > mx || a[xx] > mx) {
        ans += cnt / k * min(x, k * y);
      } else {
        cnt -= k;
        ans += x;
        ans += cnt / k * min(x, k * y);
      }
      ans += cnt % k * y;
    } else {
      if (a[l] < mx && mx > a[xx]) {
        puts("-1");
        return 0;
      }
      ans += cnt * y;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```