#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int a[N], b[N], vis[N][2], vid, n, m;
long long dp[N][2], x, k, y;
long long solve(int len, bool one) {
  if (len < 0) return 1e16;
  if (len == 0) return 0;
  long long &ret = dp[len][one];
  if (vis[len][one] == vid) return ret;
  vis[len][one] = vid;
  ret = 1e16;
  if (len >= k) ret = min(ret, solve(len - k, one) + x);
  if (len > 1 or (len == 1 and one)) ret = min(ret, solve(len - 1, one) + y);
  return ret;
}
int main() {
  ++vid;
  scanf("%d%d%lld%lld%lld", &n, &m, &x, &k, &y);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int i = 0; i < m; i++) scanf("%d", b + i);
  int ind = 0;
  vector<int> pos, val;
  pos.emplace_back(-1);
  val.emplace_back(0);
  for (int i = 0; i < n and ind < m; i++)
    if (b[ind] == a[i]) pos.emplace_back(i), val.emplace_back(b[ind++]);
  pos.emplace_back(n);
  val.emplace_back(0);
  if (ind != m) return puts("-1"), 0;
  long long ans = 0;
  for (int i = 1; i < pos.size(); i++) {
    int mx = 0, l = pos[i - 1] + 1, r = pos[i], mxB = 0;
    for (int j = l; j < r; j++) mx = max(mx, a[j]);
    mxB = max(val[i - 1], val[i]);
    long long temp = solve(pos[i] - pos[i - 1] - 1, (mxB > mx));
    if (temp == 1e16) return puts("-1"), 0;
    ans += temp;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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>
using namespace std;
const int N = 2e5 + 5;
int a[N], b[N], vis[N][2], vid, n, m;
long long dp[N][2], x, k, y;
long long solve(int len, bool one) {
  if (len < 0) return 1e16;
  if (len == 0) return 0;
  long long &ret = dp[len][one];
  if (vis[len][one] == vid) return ret;
  vis[len][one] = vid;
  ret = 1e16;
  if (len >= k) ret = min(ret, solve(len - k, one) + x);
  if (len > 1 or (len == 1 and one)) ret = min(ret, solve(len - 1, one) + y);
  return ret;
}
int main() {
  ++vid;
  scanf("%d%d%lld%lld%lld", &n, &m, &x, &k, &y);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int i = 0; i < m; i++) scanf("%d", b + i);
  int ind = 0;
  vector<int> pos, val;
  pos.emplace_back(-1);
  val.emplace_back(0);
  for (int i = 0; i < n and ind < m; i++)
    if (b[ind] == a[i]) pos.emplace_back(i), val.emplace_back(b[ind++]);
  pos.emplace_back(n);
  val.emplace_back(0);
  if (ind != m) return puts("-1"), 0;
  long long ans = 0;
  for (int i = 1; i < pos.size(); i++) {
    int mx = 0, l = pos[i - 1] + 1, r = pos[i], mxB = 0;
    for (int j = l; j < r; j++) mx = max(mx, a[j]);
    mxB = max(val[i - 1], val[i]);
    long long temp = solve(pos[i] - pos[i - 1] - 1, (mxB > mx));
    if (temp == 1e16) return puts("-1"), 0;
    ans += temp;
  }
  printf("%lld\n", ans);
  return 0;
}

```