Vasya takes part in the orienteering competition. There are n checkpoints located along the line at coordinates x1, x2, …, xn. Vasya starts at the point with coordinate a. His goal is to visit at least n - 1 checkpoint in order to finish the competition. Participant are allowed to visit checkpoints in arbitrary order.
Vasya wants to pick such checkpoints and the order of visiting them that the total distance travelled is minimized. He asks you to calculate this minimum possible value.
Input
The first line of the input contains two integers n and a (1 ≤ n ≤ 100 000, - 1 000 000 ≤ a ≤ 1 000 000) — the number of checkpoints and Vasya’s starting position respectively.
The second line contains n integers x1, x2, …, xn ( - 1 000 000 ≤ xi ≤ 1 000 000) — coordinates of the checkpoints.
Output
Print one integer — the minimum distance Vasya has to travel in order to visit at least n - 1 checkpoint.
ExamplesInput
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3101712
Output
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7
Input
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2011–10
Output
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10
Input
Copy
5000100000
Output
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0
Note
In the first sample Vasya has to visit at least two checkpoints. The optimal way to achieve this is the walk to the third checkpoints (distance is 12 - 10 = 2) and then proceed to the second one (distance is 12 - 7 = 5). The total distance is equal to 2 + 5 = 7.
In the second sample it’s enough to visit only one checkpoint so Vasya should just walk to the point - 10.
4种情况求一下min即可；
#include<iostream>#include<cstdio>#include<algorithm>#include<cstdlib>#include<cstring>#include<string>#include<cmath>#include<map>#include<set>#include<vector>#include<queue>#include<bitset>#include<ctime>#include<deque>#include<stack>#include<functional>#include<sstream>//#include<cctype>//#pragma GCC optimize(2)usingnamespace std;#define maxn 1000005#define inf 0x3f3f3f3f//#define INF 1e18#define rdint(x) scanf(“%d”,&x)#define rdllt(x) scanf(“%lld”,&x)#define rdult(x) scanf(“%lu”,&x)#define rdlf(x) scanf(“%lf”,&x)#define rdstr(x) scanf(“%s”,x)typedeflonglong ll;typedefunsignedlonglong ull;typedefunsignedint U;#define ms(x) memset((x),0,sizeof(x))constlonglongint mod =1e9+7;#defineMod1000000000#define sq(x)(x)*(x)#define eps 1e-3typedefpair<int,int> pii;#define pi acos(-1.0)constint N =1005;#define REP(i,n)for(int i=0;i<(n);i++)typedef pair<int,int> pii;inline ll rd(){ ll x =0;char c = getchar();bool f =false;while(!isdigit(c)){if(c ==–) f =true; c = getchar();}while(isdigit(c)){ x =(x <<1)+(x <<3)+(c ^48); c = getchar();}return f ?–x : x;}ll gcd(ll a, ll b){return b ==0? a : gcd(b, a%b);}ll sqr(ll x){return x * x;}/*ll ans;ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y);ll t = x; x = y; y = t – a / b * y; return ans;}*/int n, a;ll x[maxn];int main(){//ios::sync_with_stdio(0); rdint(n); rdint(a);for(int i =1; i <= n; i++)rdllt(x[i]);ll sum=0; sort(x +1, x +1+ n);if(n ==1)cout <<0<< endl;else{ sum = inf; sum = min(sum, abs(a – x[1])+ x[n –1]– x[1]); sum = min(sum, abs(a – x[n –1])+ x[n –1]– x[1]); sum = min(sum, abs(a – x[n])+ x[n]– x[2]); sum = min(sum, abs(a – x[2])+ x[n]– x[2]); cout << sum << endl;}return0;}
