There are n people and k keys on a straight line. Every person wants to get to the office which is located on the line as well. To do that, he needs to reach some point with a key, take the key and then go to the office. Once a key is taken by somebody, it couldn‘t be taken by anybody else.

You are to determine the minimum time needed for all n people to get to the office with keys. Assume that people move a unit distance per 1 second. If two people reach a key at the same time, only one of them can take the key. A person can pass through a point with a key without taking it.

The first line contains three integers n, k and p (1 ≤ n ≤ 1 000, n ≤ k ≤ 2 000, 1 ≤ p ≤ 109) — the number of people, the number of keys and the office location.

The second line contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ 109) — positions in which people are located initially. The positions are given in arbitrary order.

The third line contains k distinct integers b1, b2, ..., bk (1 ≤ bj ≤ 109) — positions of the keys. The positions are given in arbitrary order.

Note that there can‘t be more than one person or more than one key in the same point. A person and a key can be located in the same point.

Print the minimum time (in seconds) needed for all n to reach the office with keys.

In the first example the person located at point 20 should take the key located at point 40 and go with it to the office located at point 50. He spends 30 seconds. The person located at point 100 can take the key located at point 80 and go to the office with it. He spends 50 seconds. Thus, after 50 seconds everybody is in office with keys.

  1 /**  2  * Codeforces  3  * Problem#831D  4  * Accepted  5  * Time:15ms  6  * Memory:2100k  7  */   8 #include <iostream>  9 #include <cstdio> 10 #include <ctime> 11 #include <cmath> 12 #include <cctype> 13 #include <cstring> 14 #include <cstdlib> 15 #include <fstream> 16 #include <sstream> 17 #include <algorithm> 18 #include <map> 19 #include <set> 20 #include <stack> 21 #include <queue> 22 #include <vector> 23 #include <stack> 24 #include <cassert> 25 #ifndef WIN32 26 #define Auto "%lld" 27 #else 28 #define Auto "%I64d" 29 #endif 30 using namespace std; 31 typedef bool boolean; 32 const signed int inf = (signed)((1u << 31) - 1); 33 const signed long long llf = (signed long long)((1ull << 61) - 1); 34 const double eps = 1e-6; 35 const int binary_limit = 128; 36 #define smin(a, b) a = min(a, b) 37 #define smax(a, b) a = max(a, b) 38 #define max3(a, b, c) max(a, max(b, c)) 39 #define min3(a, b, c) min(a, min(b, c)) 40 template<typename T> 41 inline boolean readInteger(T& u){ 42     char x; 43     int aFlag = 1; 44     while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1); 45     if(x == -1) { 46         ungetc(x, stdin);     47         return false; 48     } 49     if(x == ‘-‘){ 50         x = getchar(); 51         aFlag = -1; 52     } 53     for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘); 54     ungetc(x, stdin); 55     u *= aFlag; 56     return true; 57 } 58  59 int n, m, p; 60 int *ps, *ks;  61  62 inline void init() { 63     readInteger(n); 64     readInteger(m); 65     readInteger(p); 66     ps = new int[(n + 1)]; 67     ks = new int[(m + 1)]; 68     for(int i = 1; i <= n; i++) 69         readInteger(ps[i]); 70     for(int i = 1; i <= m; i++) 71         readInteger(ks[i]); 72 } 73  74 pair<int, int>* ls; 75 boolean check(int mid) { 76     for(int i = 1; i <= n; i++) { 77         int dis = abs(ps[i] - p); 78         if(dis > mid)    return false; 79         dis = (mid - dis) >> 1; 80         int l = ps[i], r = p; 81         if(l > r)    swap(l, r); 82         l -= dis, r += dis; 83         ls[i] = pair<int, int>(l, r); 84     } 85 //    cout << mid << " " << ls[1].first << " " << ls[1].second << endl; 86     sort(ls + 1, ls + n + 1); 87     int i, fin; 88     for(i = 1, fin = 1; i <= n; i++, fin++) { 89         while(fin <= m && ks[fin] < ls[i].first)    fin++; 90         if((ls[i].second < ks[fin]) || fin > m) 91             return false; 92     } 93     return true; 94 }  95  96 inline void solve() { 97     sort(ks + 1, ks + m + 1); 98 //    sort(ps + 1, ps + n + 1); 99     ls = new pair<int, int>[n + 1];100     int l = 0, r = 2e9;101     while(l <= r) {102         int mid = l + ((r - l) >> 1);103         if(check(mid))    r = mid - 1;104         else l = mid + 1;105     }106     printf("%d", r + 1);107 }108 109 int main() {110     init();111     solve();112     return 0;113 }