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 PB push_back
#define MP make_pair
#define INS insert
#define LB lower_bound
#define UB upper_bound
#define pii pair <int,int>
#define pll pair <long long, long long>
#define si pair<string, int>
#define is pair<int, string>
#define X first
#define Y second
#define _ << " " <<
#define sz(x) (int)x.size()
#define all(a) (a).begin(),(a).end()
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORD(i, a, b) for (int i = (a); i > (b); --i)
#define FORR(i, l, r) for (int i = (l); i <= (r); ++i)
#define FORP(i, a, b) for ((i) = (a); (i) < (b); ++i)
#define FORA(i, x) for (auto &i : x)
#define REP(i, n) FOR(i, 0, n)
#define BITS(x) __builtin_popcount(x)
#define MSET memset
#define MCPY memcpy
#define SQ(a) (a) * (a)
#define TRACE(x) cout << #x " = " << (x) << '\n';
#define YES cout << "YES\n"
#define NO cout << "NO\n"

typedef long long ll;
typedef long double ld;
typedef vector <int> vi;
typedef vector <pii> vpi;
typedef vector <ll> vll;
typedef vector <pll> vpl;
typedef vector <double> vd;
typedef vector <ld> vld;
typedef vector<si> vsi;
typedef vector<is> vis;
typedef vector<string> vs;
//((float) t)/CLOCKS_PER_SEC

const int MOD = 1e9 + 7;
const double PI = acos(-1);
const int INF = 1e9 + 10;
const ll INFL = 1e18 + 10;
const int ABC = 30;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
const int dox[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int doy[] = {0, 0, -1, 1, -1, 1, -1, 1};

inline int sum(int a, int b){
	if (a + b < 0)
		return (a + b + MOD) % MOD;
	return (a + b) % MOD;
}

inline void add(int &a, int b){
	a = sum(a, b);
}

inline int mul(int a, int b){
	return (ll)a * (ll)b % MOD;
}

inline int sub(int a, int b){
	return (a - b + MOD) % MOD;
}

inline int fast_pot(ll pot, int n){
	ll ret = 1;
	while (n){
		if (n & 1)
			ret = (ret * pot) % MOD;
			pot = (pot * pot) % MOD;
			n >>= 1;
		}
	return ret;
}

inline int divide(int a, int b){
	return mul(a, fast_pot(b, MOD - 2));
}

ll lcm(ll a, ll b){
	return abs(a * b) / __gcd(a, b);
}

inline double ccw(pii A, pii B, pii C){
	return (A.X * B.Y) - (A.Y * B.X) + (B.X * C.Y) - (B.Y * C.X) + (C.X * A.Y) - (C.Y * A.X);
}

inline int CCW(pii A, pii B, pii C){
	double val = ccw(A, B, C);
	double eps = max(max(abs(A.X), abs(A.Y)), max(max(abs(B.X), abs(B.Y)), max(abs(C.X), abs(C.Y)))) / 1e9;
	if (val <= -eps)
		return -1;
	if (val >= eps)
		return 1;
	return 0;
}

void to_upper(string &x){
	REP(i, sz(x))
		x[i] = toupper(x[i]);
}

void to_lower(string &x){
	REP(i, sz(x))
		x[i] = tolower(x[i]);
}

string its(ll x){
	if (x == 0)
		return "0";
	string ret = "";
	while (x > 0){
		ret += (x % 10) + '0';
		x /= 10;
	}
	reverse(all(ret));
	return ret;
}

ll sti(string s){
	ll ret = 0;
	REP(i, sz(s)){
		ret *= 10;
		ret += (s[i] - '0');
	}
	return ret;
}

const int N = 2e5 + 10;



int main () {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	int k, n; cin >> k >> n;
	vi v(n);
	REP(i, n)
		cin >> v[i];

	REP(i, n)
		v.PB(v[i] + k);

	ll sol = INF, tmp = 0;
	FOR(i, 1, n)
		tmp += v[i] - v[i - 1];
	sol = tmp;
	FOR(l, 1, n){
		tmp -= v[l] - v[l - 1];
		tmp += v[l + n - 1] - v[l + n - 2];
		sol = min(sol, tmp);
	}

	cout << sol << '\n';

	return 0;
}