#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};
const int maxn = 1e5 + 10;
const int maxx = 600005;
const double EPS = 1e-8;
const double eps = 1e-8;
const int mod = 1e9 + 7;
template <class T>
inline T min(T a, T b, T c) {
  return min(min(a, b), c);
}
template <class T>
inline T max(T a, T b, T c) {
  return max(max(a, b), c);
}
template <class T>
inline T min(T a, T b, T c, T d) {
  return min(min(a, b), min(c, d));
}
template <class T>
inline T max(T a, T b, T c, T d) {
  return max(max(a, b), max(c, d));
}
template <class T>
inline bool scan_d(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) {
    return 0;
  }
  while (c != '-' && (c < '0' || c > '9')) {
    c = getchar();
  }
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') {
    ret = ret * 10 + (c - '0');
  }
  ret *= sgn;
  return 1;
}
inline bool scan_lf(double &num) {
  char in;
  double Dec = 0.1;
  bool IsN = false, IsD = false;
  in = getchar();
  if (in == EOF) return false;
  while (in != '-' && in != '.' && (in < '0' || in > '9')) in = getchar();
  if (in == '-') {
    IsN = true;
    num = 0;
  } else if (in == '.') {
    IsD = true;
    num = 0;
  } else
    num = in - '0';
  if (!IsD) {
    while (in = getchar(), in >= '0' && in <= '9') {
      num *= 10;
      num += in - '0';
    }
  }
  if (in != '.') {
    if (IsN) num = -num;
    return true;
  } else {
    while (in = getchar(), in >= '0' && in <= '9') {
      num += Dec * (in - '0');
      Dec *= 0.1;
    }
  }
  if (IsN) num = -num;
  return true;
}
void Out(long long a) {
  if (a < 0) {
    putchar('-');
    a = -a;
  }
  if (a >= 10) Out(a / 10);
  putchar(a % 10 + '0');
}
void print(long long a) { Out(a), puts(""); }
long long a[200005], b[200005];
int main() {
  int n, m, ta, tb, k;
  scanf("%d%d%d%d%d", &n, &m, &ta, &tb, &k);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i < m; i++) scanf("%lld", &b[i]);
  if (k >= n || k >= m)
    puts("-1");
  else {
    long long ans = -1;
    long long t;
    for (int i = 0; i <= k; i++) {
      t = (long long)(a[i] + ta);
      int p1 = lower_bound(b, b + m, t) - b;
      if ((long long)(p1 + k - i) < m)
        ans = max(ans, b[(long long)p1 + k - i] + tb);
      else {
        puts("-1");
        return 0;
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

Arkady bought an air ticket from a city A to a city C. Unfortunately, there are no direct flights, but there are a lot of flights from A to a city B, and from B to C.

There are n flights from A to B, they depart at time moments a_1, a_2, a_3, ..., a_n and arrive at B t_a moments later.

There are m flights from B to C, they depart at time moments b_1, b_2, b_3, ..., b_m and arrive at C t_b moments later.

The connection time is negligible, so one can use the i-th flight from A to B and the j-th flight from B to C if and only if b_j ≥ a_i + t_a.

You can cancel at most k flights. If you cancel a flight, Arkady can not use it.

Arkady wants to be in C as early as possible, while you want him to be in C as late as possible. Find the earliest time Arkady can arrive at C, if you optimally cancel k flights. If you can cancel k or less flights in such a way that it is not possible to reach C at all, print -1.

Input

The first line contains five integers n, m, t_a, t_b and k (1 ≤ n, m ≤ 2 ⋅ 10^5, 1 ≤ k ≤ n + m, 1 ≤ t_a, t_b ≤ 10^9) — the number of flights from A to B, the number of flights from B to C, the flight time from A to B, the flight time from B to C and the number of flights you can cancel, respectively.

The second line contains n distinct integers in increasing order a_1, a_2, a_3, ..., a_n (1 ≤ a_1 < a_2 < … < a_n ≤ 10^9) — the times the flights from A to B depart.

The third line contains m distinct integers in increasing order b_1, b_2, b_3, ..., b_m (1 ≤ b_1 < b_2 < … < b_m ≤ 10^9) — the times the flights from B to C depart.

Output

If you can cancel k or less flights in such a way that it is not possible to reach C at all, print -1.

Otherwise print the earliest time Arkady can arrive at C if you cancel k flights in such a way that maximizes this time.

Examples

Input


4 5 1 1 2
1 3 5 7
1 2 3 9 10


Output


11


Input


2 2 4 4 2
1 10
10 20


Output


-1


Input


4 3 2 3 1
1 999999998 999999999 1000000000
3 4 1000000000


Output


1000000003

Note

Consider the first example. The flights from A to B depart at time moments 1, 3, 5, and 7 and arrive at B at time moments 2, 4, 6, 8, respectively. The flights from B to C depart at time moments 1, 2, 3, 9, and 10 and arrive at C at time moments 2, 3, 4, 10, 11, respectively. You can cancel at most two flights. The optimal solution is to cancel the first flight from A to B and the fourth flight from B to C. This way Arkady has to take the second flight from A to B, arrive at B at time moment 4, and take the last flight from B to C arriving at C at time moment 11.

In the second example you can simply cancel all flights from A to B and you're done.

In the third example you can cancel only one flight, and the optimal solution is to cancel the first flight from A to B. Note that there is still just enough time to catch the last flight from B to C.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};
const int maxn = 1e5 + 10;
const int maxx = 600005;
const double EPS = 1e-8;
const double eps = 1e-8;
const int mod = 1e9 + 7;
template <class T>
inline T min(T a, T b, T c) {
  return min(min(a, b), c);
}
template <class T>
inline T max(T a, T b, T c) {
  return max(max(a, b), c);
}
template <class T>
inline T min(T a, T b, T c, T d) {
  return min(min(a, b), min(c, d));
}
template <class T>
inline T max(T a, T b, T c, T d) {
  return max(max(a, b), max(c, d));
}
template <class T>
inline bool scan_d(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) {
    return 0;
  }
  while (c != '-' && (c < '0' || c > '9')) {
    c = getchar();
  }
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') {
    ret = ret * 10 + (c - '0');
  }
  ret *= sgn;
  return 1;
}
inline bool scan_lf(double &num) {
  char in;
  double Dec = 0.1;
  bool IsN = false, IsD = false;
  in = getchar();
  if (in == EOF) return false;
  while (in != '-' && in != '.' && (in < '0' || in > '9')) in = getchar();
  if (in == '-') {
    IsN = true;
    num = 0;
  } else if (in == '.') {
    IsD = true;
    num = 0;
  } else
    num = in - '0';
  if (!IsD) {
    while (in = getchar(), in >= '0' && in <= '9') {
      num *= 10;
      num += in - '0';
    }
  }
  if (in != '.') {
    if (IsN) num = -num;
    return true;
  } else {
    while (in = getchar(), in >= '0' && in <= '9') {
      num += Dec * (in - '0');
      Dec *= 0.1;
    }
  }
  if (IsN) num = -num;
  return true;
}
void Out(long long a) {
  if (a < 0) {
    putchar('-');
    a = -a;
  }
  if (a >= 10) Out(a / 10);
  putchar(a % 10 + '0');
}
void print(long long a) { Out(a), puts(""); }
long long a[200005], b[200005];
int main() {
  int n, m, ta, tb, k;
  scanf("%d%d%d%d%d", &n, &m, &ta, &tb, &k);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i < m; i++) scanf("%lld", &b[i]);
  if (k >= n || k >= m)
    puts("-1");
  else {
    long long ans = -1;
    long long t;
    for (int i = 0; i <= k; i++) {
      t = (long long)(a[i] + ta);
      int p1 = lower_bound(b, b + m, t) - b;
      if ((long long)(p1 + k - i) < m)
        ans = max(ans, b[(long long)p1 + k - i] + tb);
      else {
        puts("-1");
        return 0;
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```