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)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repo(i,n) for(int i = 1; i < (int)(n); i++)
#define pb push_back
#define np next_permutation
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define mod 1000000007
#define pi acos(-1.0)
ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }
//cout << fixed << setprecision (20);   小数点以下2０桁まで
//intの最大値2147483647 ≒ 2×10^9
//long longの最大値9223372036854775807 ≒ 9×10^18
//'0'+=16;        で大文字に
//'大文字'+=32;    で小文字に
//s[i]-'0'でchar文字→int数に;
//string s = to_string(int数);
//int n = stoi(string文字列)
//実行時間制約2秒では２×10^8回くらいまで計算できる


int main() {
  int k,n;
  cin >> k >> n;
  
  int x=0;
  vector<int> p(n);
  cin>>p[0];
  
  repo(i,n){
  cin>>p[i];
  x=max(x,p[i]-p[i-1]);
  }
  
  x=max(x,k-(p[n-1]-p[0]));
  
  cout << k-x << endl;
    
}