#include <bits/stdc++.h>
using namespace std;
long long readint() {
  long long x = 0, s = 1;
  char c;
  while ((c = getchar()), c < '0' || c > '9') {
    if (c == '-') s = -1;
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return s * x;
}
void print(long long x) {
  char c[20];
  if (!x) {
    putchar('0');
    return;
  }
  if (x < 0) putchar('-'), x = -x;
  int i = 0;
  while (x > 0) c[++i] = x % 10, x /= 10;
  while (i) putchar(c[i--] + 48);
}
void scan(long long& x) { x = readint(); }
template <typename T, typename... A>
void print(T& x, A&... a) {
  print(x);
  putchar(' ');
  print(a...);
}
template <typename T, typename... A>
void scan(T& x, A&... a) {
  scan(x);
  scan(a...);
}
long long n, m, k, x, y;
long long A[200200], B[200200];
long long C[200200], D[200200];
long long cost(long long l, long long r) {
  long long has = max((l == 0 ? -2000000007 : A[l - 1]),
                      (r == n - 1 ? -2000000007 : A[r + 1])) <
                  *max_element(A + l, A + r + 1);
  long long pre = -2000000007;
  long long len = r - l + 1;
  long long ret = 0;
  if (has) {
    if (len < k) {
      printf("-1\n");
      exit(0);
    }
    if (k * y <= x) {
      len -= k;
      return len * y + x;
    } else {
      return (len % k) * y + len / k * x;
    }
  } else {
    if (len < k || k * y <= x) {
      return len * y;
    } else {
      return (len % k) * y + len / k * x;
    }
  }
  return ret;
}
int32_t main() {
  scan(n, m);
  scan(x, k, y);
  memset(C, -1, sizeof C);
  memset(D, -1, sizeof D);
  for (long long i = 0; i < n; i++) C[A[i] = readint()] = i;
  for (long long i = 0; i < m; i++) D[B[i] = readint()] = i;
  {};
  long long pre = -2000000007;
  for (long long i = 0; i < m; i++) {
    if (C[B[i]] == -1)
      return printf("-1\n"), 0;
    else {
      if (C[B[i]] < pre)
        return printf("-1\n"), 0;
      else
        pre = C[B[i]];
    }
  }
  {};
  long long ans = 0;
  for (long long i = 0; i < n;) {
    if (D[A[i]] != -1) {
      i++;
      continue;
    }
    long long j = i;
    while (j < n && D[A[j]] == -1) j++;
    ans += cost(i, j - 1);
    i = j;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long readint() {
  long long x = 0, s = 1;
  char c;
  while ((c = getchar()), c < '0' || c > '9') {
    if (c == '-') s = -1;
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return s * x;
}
void print(long long x) {
  char c[20];
  if (!x) {
    putchar('0');
    return;
  }
  if (x < 0) putchar('-'), x = -x;
  int i = 0;
  while (x > 0) c[++i] = x % 10, x /= 10;
  while (i) putchar(c[i--] + 48);
}
void scan(long long& x) { x = readint(); }
template <typename T, typename... A>
void print(T& x, A&... a) {
  print(x);
  putchar(' ');
  print(a...);
}
template <typename T, typename... A>
void scan(T& x, A&... a) {
  scan(x);
  scan(a...);
}
long long n, m, k, x, y;
long long A[200200], B[200200];
long long C[200200], D[200200];
long long cost(long long l, long long r) {
  long long has = max((l == 0 ? -2000000007 : A[l - 1]),
                      (r == n - 1 ? -2000000007 : A[r + 1])) <
                  *max_element(A + l, A + r + 1);
  long long pre = -2000000007;
  long long len = r - l + 1;
  long long ret = 0;
  if (has) {
    if (len < k) {
      printf("-1\n");
      exit(0);
    }
    if (k * y <= x) {
      len -= k;
      return len * y + x;
    } else {
      return (len % k) * y + len / k * x;
    }
  } else {
    if (len < k || k * y <= x) {
      return len * y;
    } else {
      return (len % k) * y + len / k * x;
    }
  }
  return ret;
}
int32_t main() {
  scan(n, m);
  scan(x, k, y);
  memset(C, -1, sizeof C);
  memset(D, -1, sizeof D);
  for (long long i = 0; i < n; i++) C[A[i] = readint()] = i;
  for (long long i = 0; i < m; i++) D[B[i] = readint()] = i;
  {};
  long long pre = -2000000007;
  for (long long i = 0; i < m; i++) {
    if (C[B[i]] == -1)
      return printf("-1\n"), 0;
    else {
      if (C[B[i]] < pre)
        return printf("-1\n"), 0;
      else
        pre = C[B[i]];
    }
  }
  {};
  long long ans = 0;
  for (long long i = 0; i < n;) {
    if (D[A[i]] != -1) {
      i++;
      continue;
    }
    long long j = i;
    while (j < n && D[A[j]] == -1) j++;
    ans += cost(i, j - 1);
    i = j;
  }
  cout << ans << endl;
  return 0;
}

```