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)

//#pragma GCC optimize ("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/rope>
#define pb push_back
#define F first
#define S second
#define ins insert
#define mp make_pair
#define fo(i, n1, n, x) for(int i = n1; i <= n; i += x)
#define foo(i, n, n1, x) for(int i = n; i >= n1; i -= x)
#define bit __builtin_popcount
#define md (l + ((r - l) / 2))
#define all(x) x.begin(),x.end()
#define eb emplace_back
#define ub upper_bound
#define lb lower_bound
#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define file(s) if (fopen(s".in", "r")) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)

using namespace std;
//using namespace __gnu_pbds;
//using namespace __gnu_cxx;


using ll = long long;

//#define int ll

using pii = pair<int,int>;
using pll = pair<ll,ll>;
using vi = vector<int>;



const int N = 2e5 + 11, mod = 1e9 + 7, mod2 = 998244353;
const int MAX = 1e5 + 11;
const int INF1 = 2e9 + 11;
const ll INF2 = 2e18 + 11;
const double INF3 = 1e8 + 11;
const int base = 500;
const int P = 31;
const int dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
const int dy[] = {0, 0, 1, -1, 1, -1, 1, -1};
const double EPS = 1e-4;
const double PI = acos(-1.0);


//template<typename T> using ordered_set = tree <T, null_type, less <T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T1, typename T2> inline void chmin(T1 &a, T2 b) { if (a > b) a = b; }
template<typename T1, typename T2> inline void chmax(T1 &a, T2 b) { if (a < b) a = b; }


mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, k, a[N], ans = INF1;
main() {
  file("threesum");
  ios;
  cin >> k >> n;
  fo(i, 1, n, 1) cin >> a[i];
  sort(a + 1, a + n + 1);
  chmin(ans, a[n]-a[1]);
  fo(i, 1, n, 1) {
    chmin(ans, k - a[i] + a[i - 1]);
    chmin(ans, a[i] + k - a[i + 1]);
  }
  cout << ans;
  return 0;
}
/**

*/