#include <bits/stdc++.h>
using namespace std;
using pll = pair<long long, long long>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long gcd_helper(long long a, long long b) {
  if (a % b == 0)
    return b;
  else
    return gcd_helper(b, a % b);
}
long long gcd(long long a, long long b) {
  if (a < 0) a *= -1;
  if (b < 0) b *= -1;
  if (a < b) swap(a, b);
  if (b == 0) return a;
  return gcd_helper(a, b);
}
long long powr(long long a, long long b, long long M) {
  if (b == 0) return 1;
  if (a == 0) return 0;
  long long ret = 1;
  while (b) {
    if (b & 1) {
      ret *= a;
      ret %= M;
    }
    a = (a * a) % M;
    b /= 2;
  }
  return ret;
}
long long find(long long x, vector<long long> &p) {
  if (p[x] != x) p[x] = find(p[x], p);
  return p[x];
}
void unite(long long x, long long y, vector<long long> &p,
           vector<long long> &sz) {
  long long px = find(x, p);
  long long py = find(y, p);
  if (sz[px] < sz[py]) {
    swap(px, py);
    swap(x, y);
  }
  p[py] = px;
  sz[px] += sz[py];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  vector<long long> a(n + 2, 0), b(m + 2, 0);
  a[n + 1] = b[m + 1] = -1;
  for (long long i = 0; i < (n); i++) cin >> a[i + 1];
  for (long long i = 0; i < (m); i++) cin >> b[i + 1];
  long long j = 0, k2 = 0, l, mx, ans = 0, q, r, len;
  bool ok = true, firecheap = false;
  if (x < y * k) firecheap = true;
  for (long long i = (1); i < (m + 2); i += 1) {
    while (k2 < n + 2 && a[k2] != b[i]) k2++;
    if (k2 == n + 2) {
      ok = false;
      break;
    }
    if (k2 == j + 1) {
      j = k2;
      continue;
    }
    l = k2 - j - 1;
    mx = max(b[i - 1], b[i]);
    vector<long long> temp;
    for (long long z = (j + 1); z < (k2); z += 1) {
      if (a[z] > mx) temp.push_back(z);
    }
    if (l < k && temp.size()) {
      ok = false;
      break;
    }
    r = l % k;
    q = l / k;
    ans += r * y;
    l -= r;
    if (x <= y * k) {
      ans += q * x;
    } else if (temp.empty()) {
      ans += l * y;
    } else {
      ans += x + (l - k) * y;
    }
    j = k2;
  }
  if (!ok)
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}

### Prompt

Generate 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;
using pll = pair<long long, long long>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long gcd_helper(long long a, long long b) {
  if (a % b == 0)
    return b;
  else
    return gcd_helper(b, a % b);
}
long long gcd(long long a, long long b) {
  if (a < 0) a *= -1;
  if (b < 0) b *= -1;
  if (a < b) swap(a, b);
  if (b == 0) return a;
  return gcd_helper(a, b);
}
long long powr(long long a, long long b, long long M) {
  if (b == 0) return 1;
  if (a == 0) return 0;
  long long ret = 1;
  while (b) {
    if (b & 1) {
      ret *= a;
      ret %= M;
    }
    a = (a * a) % M;
    b /= 2;
  }
  return ret;
}
long long find(long long x, vector<long long> &p) {
  if (p[x] != x) p[x] = find(p[x], p);
  return p[x];
}
void unite(long long x, long long y, vector<long long> &p,
           vector<long long> &sz) {
  long long px = find(x, p);
  long long py = find(y, p);
  if (sz[px] < sz[py]) {
    swap(px, py);
    swap(x, y);
  }
  p[py] = px;
  sz[px] += sz[py];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  vector<long long> a(n + 2, 0), b(m + 2, 0);
  a[n + 1] = b[m + 1] = -1;
  for (long long i = 0; i < (n); i++) cin >> a[i + 1];
  for (long long i = 0; i < (m); i++) cin >> b[i + 1];
  long long j = 0, k2 = 0, l, mx, ans = 0, q, r, len;
  bool ok = true, firecheap = false;
  if (x < y * k) firecheap = true;
  for (long long i = (1); i < (m + 2); i += 1) {
    while (k2 < n + 2 && a[k2] != b[i]) k2++;
    if (k2 == n + 2) {
      ok = false;
      break;
    }
    if (k2 == j + 1) {
      j = k2;
      continue;
    }
    l = k2 - j - 1;
    mx = max(b[i - 1], b[i]);
    vector<long long> temp;
    for (long long z = (j + 1); z < (k2); z += 1) {
      if (a[z] > mx) temp.push_back(z);
    }
    if (l < k && temp.size()) {
      ok = false;
      break;
    }
    r = l % k;
    q = l / k;
    ans += r * y;
    l -= r;
    if (x <= y * k) {
      ans += q * x;
    } else if (temp.empty()) {
      ans += l * y;
    } else {
      ans += x + (l - k) * y;
    }
    j = k2;
  }
  if (!ok)
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}

```