Problem:
Score : 300 points

Problem Statement

There is a circular pond with a perimeter of K meters, and N houses around them.

The i-th house is built at a distance of A_i meters from the northmost point of the pond, measured clockwise around the pond.

When traveling between these houses, you can only go around the pond.

Find the minimum distance that needs to be traveled when you start at one of the houses and visit all the N houses.

Constraints

2 \leq K \leq 10^6

2 \leq N \leq 2 \times 10^5

0 \leq A_1 < ... < A_N < K

All values in input are integers.

Input

Input is given from Standard Input in the following format:

K N
A_1 A_2 ... A_N

Output

Print the minimum distance that needs to be traveled when you start at one of the houses and visit all the N houses.

Sample Input 1

20 3
5 10 15

Sample Output 1

10

If you start at the 1-st house and go to the 2-nd and 3-rd houses in this order, the total distance traveled will be 10.

Sample Input 2

20 3
0 5 15

Sample Output 2

10

If you start at the 2-nd house and go to the 1-st and 3-rd houses in this order, the total distance traveled will be 10.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"
#define SIZE          100005
#define REM           1000000007
#define PI            2*acos(0.0)
#define L             long
#define LL            long long
#define UL            unsigned long
#define ULL           unsigned long long
#define TERMINATE     return 0
#define FAST          ios_base::sync_with_stdio(false);cin.tie(NULL)
using namespace std;
template <class T> inline T bigmod1(T p,T e,T M){
    LL ret=1;
    for(; e > 0; e >>= 1){
 
        if(e & 1) ret = (ret * p) % M;
        p = (p * p) % M;
    } return (T)ret;
}
 
template <class T> inline T gcode(T a,T b){if(b==0)return a;return gcode(b,a%b);}
template <class T> inline T modularinv(T a,T M){return bigmod1(a,M-2,M);}
 
bool SA(const pair<string,string> &a, const pair<string,string> &b){
   return a.second<b.second;
}
bool FA(const pair<string,string> &a, const pair<string,string> &b){
   return a.first<b.first;
}
bool SD(const pair<LL,LL> &a, const pair<LL,LL> &b){
   return a.second>b.second;
}
bool FD(const pair<int,int> &a, const pair<int,int> &b){
   return a.first>b.first;
}
//int dirA[8]={0,0,1,-1,1,-1,1,-1};
//int dirB[8]={1,-1,0,0,1,-1,-1,1};
//int dirA[4]={0,0,1,-1};
//int dirB[4]={1,-1,0,0};
//int kingsA[8]={0,0,1,-1,-1,1,-1,1};  //Kings Move
//int kingsB[8]={-1,1,0,0,1,1,-1,-1};  //Kings Move
//int knightsA[8]={-2,-2,-1,-1,1,1,2,2};  //Knights Move
//int knightsB[8]={-1,1,-2,2,-2,2,-1,1};  //Knights Move

int main(){
	FAST;
	LL a,b;
	cin>>a>>b;
	LL ar[b];
	for(LL i=0;i<b;i++){
		cin>>ar[i];
	}
	sort(ar,ar+b);	
	LL sum0=0;
	for(LL i=1;i<b;i++){
		sum0+=abs(ar[i]-ar[i-1]);
	}
	LL sum1=0;
	if(ar[0]==0) sum1+=0;
	else sum1+=(20-ar[0]);
	for(LL i=2;i<b;i++){
		sum1+=abs(ar[i]-ar[i-1]);
	}
	if(ar[b-1]==0) sum1+=0;
	else sum1+=abs(a-ar[b-1]);
	LL sum2=abs(ar[1]-ar[0]);
	if(ar[0]==0) sum2+=abs(a-(ar[0]+20));
	else sum2+=abs(a-ar[0]);
	//cout<<sum2<<endl;
	for(LL i=3;i<b;i++){
		sum2+=abs(ar[i]-ar[i-1]);
	}
	//cout<<sum2<<endl;
	if(ar[b-1]==0) sum2+=abs(a-(ar[b-1]+20));
	else sum2+=abs(a-ar[b-1]);
	cout<<min(sum1,min(sum2,sum0))<<endl;
	cerr<<"time elapsed:"<<clock()*1000.0/CLOCKS_PER_SEC<<"ms"<<"\n";
	return 0;
}