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>
 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define rrep(i, n) for(int i = ((int)(n)-1); i >= 0; i--)
#define all(x) (x).begin(),(x).end()
 
using namespace std;
using ll = long long;
 
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

struct edge { int to; int cost; };
 
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> P;
 
const int INF = 1e9;
const int MOD = 1e9+7;
const double EPS = 1.0e-10;
const double PI = acos(-1.0);
 
const int mx4[] = {0,1,0,-1};
const int my4[] = {1,0,-1,0};
const int mx8[] = {0,1,1,1,0,-1,-1,-1};
const int my8[] = {1,1,0,-1,-1,-1,0,1};

int main() {
	int k,n;
	cin >> k >> n;
	vi a(n);
	rep(i,n) cin >> a[i];
	sort(all(a));
	int d = k-(a[n-1]-a[0]);
	for(int i = 1; i < n; i++){
		d = max(d,a[i]-a[i-1]);
	}
	cout << k-d << endl;
	return 0;
}