B. Born This Waytime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputArkady 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 \ge 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.InputThe first line contains five integers n, m, t_a, t_b and k (1 \le n, m \le 2 \cdot 10^5, 1 \le k \le n + m, 1 \le t_a, t_b \le 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 \le a_1 < a_2 < \ldots < a_n \le 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 \le b_1 < b_2 < \ldots < b_m \le 10^9) — the times the flights from B to C depart.OutputIf 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.ExamplesInput
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
NoteConsider 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.
4 5 1 1 2
1 3 5 7
1 2 3 9 10

11 
['binary search', 'brute force', 'two pointers', '*1600']