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: Accepted)

#include <bits/stdc++.h>

using namespace std;
#define inf             0x3f3f3f3f
#define INF             2e18
#define eps             1e-9

#define FastRead        ios_base::sync_with_stdio(0);cin.tie(0)
#define fRead           freopen("input.txt","r",stdin)
#define fWrite          freopen("output.txt","w",stdout)

#define LL              long long
#define ull            unsigned long long
#define PI              acos(-1.0)
#define pb              push_back
#define mk              make_pair
#define pii             pair<int,int>
#define pLL             pair<LL,LL>
#define ff              first
#define ss              second
#define all(a)          a.begin(),a.end()
#define SQR(a)          ((a)*(a))
#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(min(a,b),min(c,d))
#define max4(a,b,c,d)   max(max(a,b),max(c,d))
#define max5(a,b,c,d,e) max(max3(a,b,c),max(d,e))
#define min5(a,b,c,d,e) min(min3(a,b,c),min(d,e))
#define vi              vector <int>
#define vL              vector <LL>
#define LB(a,x)         (lower_bound(all(a),x)-a.begin()) //  first element in the range [first,last) which does not compare less than val.
#define UB(a,x)         (upper_bound(all(a),x)-a.begin()) //  first element in the range [first,last) which compares greater than val.
#define prec(n)         fixed << setprecision(n)

#define MEM(a,x)        memset(a,x,sizeof(a))
#define SORT(v)         sort(v.begin(),v.end())
#define REV(v)          reverse(v.begin(),v.end())
#define Unique(a)       sort(all(a)),a.erase(unique(all(a)),a.end())

#define FOR(i,a,b)      for(int i=a;i<=b;i++)
#define ROF(i,a,b)      for(int i=a;i>=b;i--)
#define REP(i,b)        for(int i=0;i<b;i++)
#define IT(it,x)	    for(it=x.begin();it!=x.end();it++)
LL dis1[1000009];
LL dis2[1000009];
int main()
{
    LL k,n;
    cin>>k>>n;
    LL a[n];
    for(LL i=0; i<n; i++)
        cin>>a[i];
    dis1[0]=a[0];
    for(LL i=1; i<n; i++)
        dis1[i]=a[i]-a[i-1]+dis1[i-1];
    dis2[n-1]=k-a[n-1];
    for(LL i=n-2; i>=0; i--)
    {
        dis2[i]=a[i+1]-a[i]+dis2[i+1];
    }
    LL mx=INF;
//    for(LL i=0;i<n;i++)cout<<dis1[i]<<" ";
//    cout<<endl;
//        for(LL i=0;i<n;i++)cout<<dis2[i]<<" ";
//    cout<<endl;
    for(LL i=0; i<n; i++)
    {
        mx=min(mx,dis1[i]+dis2[i+1]);
    }
    LL p=0;
    for(LL i=1;i<n;i++)
    {
        p+=a[i]-a[i-1];
    }
    mx=min(mx,p);
    p=0;
    for(LL i=n-2;i>=0;i--)
    {
        p+=a[i+1]-a[i];
    }
    mx=min(mx,p);
    cout<<mx;



}