#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast,no-stack-protector", 2, 3)
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
template <typename T>
void read(T &a) {
  T x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  a = x * f;
}
template <typename T>
void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10) return (void)putchar(x + '0');
  write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  putchar('\n');
}
template <typename T>
void writes(T x) {
  write(x);
  putchar(' ');
}
template <typename T, typename... Args>
void read(T &t, Args &...args) {
  read(t);
  read(args...);
}
template <typename T, typename... Args>
void writes(T t, Args... args) {
  writes(t);
  writes(args...);
}
template <typename T, typename... Args>
void writeln(T t, Args... args) {
  writes(t);
  writes(args...);
  putchar('\n');
}
long long n, m, x, y, k, a[200005], b[200005], top = 1, ans, cnt, maxx;
bool vis[200005];
signed main() {
  read(n, m, x, k, y);
  for (long long i = 1; i <= n; i++) read(a[i]);
  for (long long i = 1; i <= m; i++) read(b[i]);
  for (long long i = 1; i <= n; i++)
    if (a[i] == b[top]) vis[i] = 1, top++;
  if (top < m) return puts("-1"), 0;
  long long lst = 0;
  vis[n + 1] = 1;
  for (long long i = 1; i <= n + 1; i++) {
    if (vis[i]) {
      cnt = i - lst - 1;
      if (!cnt) {
        lst = i;
        maxx = 0;
        continue;
      }
      if (cnt < k) {
        if (maxx < max(a[lst], a[i]))
          ans += y * cnt;
        else
          return puts("-1"), 0;
      } else {
        if (y * k < x) {
          if (maxx < max(a[lst], a[i]))
            ans += y * cnt;
          else
            ans += y * (cnt - k) + x;
        } else {
          ans += cnt / k * x + (cnt % k) * y;
        }
      }
      lst = i;
      maxx = 0;
    } else
      maxx = max(maxx, a[i]);
  }
  writeln(ans);
  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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast,no-stack-protector", 2, 3)
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
template <typename T>
void read(T &a) {
  T x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  a = x * f;
}
template <typename T>
void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10) return (void)putchar(x + '0');
  write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  putchar('\n');
}
template <typename T>
void writes(T x) {
  write(x);
  putchar(' ');
}
template <typename T, typename... Args>
void read(T &t, Args &...args) {
  read(t);
  read(args...);
}
template <typename T, typename... Args>
void writes(T t, Args... args) {
  writes(t);
  writes(args...);
}
template <typename T, typename... Args>
void writeln(T t, Args... args) {
  writes(t);
  writes(args...);
  putchar('\n');
}
long long n, m, x, y, k, a[200005], b[200005], top = 1, ans, cnt, maxx;
bool vis[200005];
signed main() {
  read(n, m, x, k, y);
  for (long long i = 1; i <= n; i++) read(a[i]);
  for (long long i = 1; i <= m; i++) read(b[i]);
  for (long long i = 1; i <= n; i++)
    if (a[i] == b[top]) vis[i] = 1, top++;
  if (top < m) return puts("-1"), 0;
  long long lst = 0;
  vis[n + 1] = 1;
  for (long long i = 1; i <= n + 1; i++) {
    if (vis[i]) {
      cnt = i - lst - 1;
      if (!cnt) {
        lst = i;
        maxx = 0;
        continue;
      }
      if (cnt < k) {
        if (maxx < max(a[lst], a[i]))
          ans += y * cnt;
        else
          return puts("-1"), 0;
      } else {
        if (y * k < x) {
          if (maxx < max(a[lst], a[i]))
            ans += y * cnt;
          else
            ans += y * (cnt - k) + x;
        } else {
          ans += cnt / k * x + (cnt % k) * y;
        }
      }
      lst = i;
      maxx = 0;
    } else
      maxx = max(maxx, a[i]);
  }
  writeln(ans);
  return 0;
}

```