#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5;
long long a[N], b[N];
long long x, y, k, n, m;
long long tree[N * 4];
void build(long long node, long long st, long long end) {
  if (st == end) {
    tree[node] = a[st];
    return;
  }
  long long mid = (st + end) / 2;
  build(2 * node, st, mid);
  build(2 * node + 1, mid + 1, end);
  tree[node] = max(tree[2 * node], tree[2 * node + 1]);
}
long long query(long long node, long long st, long long end, long long l,
                long long r) {
  if (st > r || end < l) return 0;
  if (st >= l && end <= r) {
    return tree[node];
  }
  long long mid = (st + end) / 2;
  return max(query(2 * node, st, mid, l, r),
             query(2 * node + 1, mid + 1, end, l, r));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  std::ios::sync_with_stdio(false);
  long long te = 1;
  cin >> n >> m >> x >> k >> y;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < m; i++) cin >> b[i];
  build(1, 0, n - 1);
  vector<long long> v;
  long long i = 0, j = 0;
  while (i < n && j < m) {
    if (a[i] == b[j]) {
      v.push_back(i);
      i++, j++;
    } else {
      i++;
    }
  }
  if (v.size() != m) {
    return cout << -1, 0;
  }
  long long ans = 0;
  long long flag = 0;
  if (v[0] != 0) {
    long long req = (v[0]);
    long long x1 = v[0];
    long long res = req / k;
    if (res == 0) {
      long long curr = query(1, 0, n - 1, 0, x1);
      if (curr == a[x1]) {
        ans += req * y;
      } else {
        flag = 1;
      }
    } else {
      long long c1 = 0, c2 = 0;
      long long curr = query(1, 0, n - 1, 0, x1);
      long long rem = req % k;
      c1 += rem * y + res * x;
      if (curr == a[x1]) {
        c2 += req * y;
      } else {
        c2 = 1e18;
      }
      long long c3 = 0;
      long long exp = (x - k * y);
      if (exp < 0) {
        c3 = (req * y) / (x - k * y);
      } else {
        c3 = 1;
      }
      c3 = (exp)*c3 + req * y;
      ans += min(c1, min(c2, c3));
      ;
    }
  }
  for (long long i = 0; i < v.size() - 1; i++) {
    long long x1 = v[i], y1 = v[i + 1];
    long long req = y1 - x1 - 1;
    if (req == 0) continue;
    long long res = req / k;
    long long vari = 0;
    if (res == 0) {
      long long curr = query(1, 0, n - 1, x1, y1);
      if (curr == a[x1] || curr == a[y1]) {
        ans += req * y;
        vari = req * y;
      } else {
        flag = 1;
        break;
      }
    } else {
      long long t1 = 0, t2 = 0;
      long long curr = query(1, 0, n - 1, x1, y1);
      long long rem = req % k;
      long long c1 = rem * y + res * x;
      long long c2 = 0;
      if (curr == a[x1] || curr == a[y1]) {
        c2 += req * y;
      } else {
        c2 = 1e18;
      }
      long long c3 = 0;
      long long exp = (x - k * y);
      if (exp < 0) {
        c3 = (req * y) / (x - k * y);
      } else {
        c3 = 1;
      }
      c3 = (exp)*c3 + req * y;
      ans += min(c1, min(c2, c3));
      ;
    }
  }
  if (v[m - 1] != n - 1) {
    long long req = (n - 1 - v[m - 1]);
    long long x1 = v[m - 1];
    long long res = req / k;
    if (res == 0) {
      long long curr = query(1, 0, n - 1, x1, n - 1);
      if (curr == a[x1]) {
        ans += req * y;
      } else {
        flag = 1;
      }
    } else {
      long long c1 = 0, c2 = 0;
      long long curr = query(1, 0, n - 1, x1, n - 1);
      long long rem = req % k;
      c1 += rem * y + res * x;
      if (curr == a[x1]) {
        c2 += req * y;
      } else {
        c2 = 1e18;
      }
      long long c3 = 0;
      long long exp = (x - k * y);
      if (exp < 0) {
        c3 = (req * y) / (x - k * y);
      } else {
        c3 = 1;
      }
      c3 = (exp)*c3 + req * y;
      ans += min(c1, min(c2, c3));
      ;
    }
  }
  if (flag) {
    return cout << -1, 0;
  }
  cout << ans << "\n";
}

### 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>
using namespace std;
const long long N = 2e5 + 5;
long long a[N], b[N];
long long x, y, k, n, m;
long long tree[N * 4];
void build(long long node, long long st, long long end) {
  if (st == end) {
    tree[node] = a[st];
    return;
  }
  long long mid = (st + end) / 2;
  build(2 * node, st, mid);
  build(2 * node + 1, mid + 1, end);
  tree[node] = max(tree[2 * node], tree[2 * node + 1]);
}
long long query(long long node, long long st, long long end, long long l,
                long long r) {
  if (st > r || end < l) return 0;
  if (st >= l && end <= r) {
    return tree[node];
  }
  long long mid = (st + end) / 2;
  return max(query(2 * node, st, mid, l, r),
             query(2 * node + 1, mid + 1, end, l, r));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  std::ios::sync_with_stdio(false);
  long long te = 1;
  cin >> n >> m >> x >> k >> y;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < m; i++) cin >> b[i];
  build(1, 0, n - 1);
  vector<long long> v;
  long long i = 0, j = 0;
  while (i < n && j < m) {
    if (a[i] == b[j]) {
      v.push_back(i);
      i++, j++;
    } else {
      i++;
    }
  }
  if (v.size() != m) {
    return cout << -1, 0;
  }
  long long ans = 0;
  long long flag = 0;
  if (v[0] != 0) {
    long long req = (v[0]);
    long long x1 = v[0];
    long long res = req / k;
    if (res == 0) {
      long long curr = query(1, 0, n - 1, 0, x1);
      if (curr == a[x1]) {
        ans += req * y;
      } else {
        flag = 1;
      }
    } else {
      long long c1 = 0, c2 = 0;
      long long curr = query(1, 0, n - 1, 0, x1);
      long long rem = req % k;
      c1 += rem * y + res * x;
      if (curr == a[x1]) {
        c2 += req * y;
      } else {
        c2 = 1e18;
      }
      long long c3 = 0;
      long long exp = (x - k * y);
      if (exp < 0) {
        c3 = (req * y) / (x - k * y);
      } else {
        c3 = 1;
      }
      c3 = (exp)*c3 + req * y;
      ans += min(c1, min(c2, c3));
      ;
    }
  }
  for (long long i = 0; i < v.size() - 1; i++) {
    long long x1 = v[i], y1 = v[i + 1];
    long long req = y1 - x1 - 1;
    if (req == 0) continue;
    long long res = req / k;
    long long vari = 0;
    if (res == 0) {
      long long curr = query(1, 0, n - 1, x1, y1);
      if (curr == a[x1] || curr == a[y1]) {
        ans += req * y;
        vari = req * y;
      } else {
        flag = 1;
        break;
      }
    } else {
      long long t1 = 0, t2 = 0;
      long long curr = query(1, 0, n - 1, x1, y1);
      long long rem = req % k;
      long long c1 = rem * y + res * x;
      long long c2 = 0;
      if (curr == a[x1] || curr == a[y1]) {
        c2 += req * y;
      } else {
        c2 = 1e18;
      }
      long long c3 = 0;
      long long exp = (x - k * y);
      if (exp < 0) {
        c3 = (req * y) / (x - k * y);
      } else {
        c3 = 1;
      }
      c3 = (exp)*c3 + req * y;
      ans += min(c1, min(c2, c3));
      ;
    }
  }
  if (v[m - 1] != n - 1) {
    long long req = (n - 1 - v[m - 1]);
    long long x1 = v[m - 1];
    long long res = req / k;
    if (res == 0) {
      long long curr = query(1, 0, n - 1, x1, n - 1);
      if (curr == a[x1]) {
        ans += req * y;
      } else {
        flag = 1;
      }
    } else {
      long long c1 = 0, c2 = 0;
      long long curr = query(1, 0, n - 1, x1, n - 1);
      long long rem = req % k;
      c1 += rem * y + res * x;
      if (curr == a[x1]) {
        c2 += req * y;
      } else {
        c2 = 1e18;
      }
      long long c3 = 0;
      long long exp = (x - k * y);
      if (exp < 0) {
        c3 = (req * y) / (x - k * y);
      } else {
        c3 = 1;
      }
      c3 = (exp)*c3 + req * y;
      ans += min(c1, min(c2, c3));
      ;
    }
  }
  if (flag) {
    return cout << -1, 0;
  }
  cout << ans << "\n";
}

```