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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

/*--------switch header--------*/
#define ALGORITHM   1
#define CASSERT     0
#define CMATH       0
#define DEQUE       0
#define FUNCTIONAL  1
#define IOMANIP     1
#define IOSTREAM    1
#define LIST        0
#define MAP         0
#define QUEUE       0
#define SET         0
#define STRING      1
#define UTILITY     1
#define VECTOR      1
#define VEC_DEBUG   1

using namespace std;

#pragma region headers
#if VEC_DEBUG
#define _GLIBCXX_DEBUG
#endif
#if ALGORITHM
#include <algorithm>
#endif
#if CASSERT
#include <cassert>
#endif
#if CMATH
#include <cmath>
#endif
#if DEQUE
#include <deque>
#endif
#if FUNCTIONAL
#include <functional>
#endif
#if IOMANIP
#include <iomanip>
#endif
#if IOSTREAM
#include <iostream>
#endif
#if LIST
#include <list>
#endif
#if MAP
#include <map>
#endif
#if QUEUE
#include <queue>
#endif
#if SET
#include <set>
#endif
#if STRING
#include <string>
#endif
#if UTILITY
#include <utility>
#endif
#if VECTOR
#include <vector>
#endif
#pragma endregion


/*----------macros----------*/

//output================
#define endl '\n'
#define fcout cout << fixed << setprecision(15)

//typedef===============
typedef long long ll;
typedef long double ld;
typedef pair<int, int> Pii;
typedef pair<ll, ll> Pll;

//const number==========
constexpr ll INF9   = (ll)1e9 + 1;
constexpr ll INF18  = (ll)1e18 + 1;
constexpr ll MOD    = 1000000007;
constexpr ld PI     = 3.141592653589793L;

//member================
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second

//others================
#define ALL(x)          (x).begin(), (x).end()
#define rALL(x)         (x).rbegin(), (x).rend()
#define REP(i, n)       for(int i = 0; (i) < (n); (i)++)
#define rREP(i, n)      for(int i = n - 1; (i) >= 0 ; (i)--)
#define FOR(i, a, b)    for(int i = (a); (i) < (b) ; (i)++)
#define CtoI(c)         int((c)) - int('0')

//functions=============
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }

//vector<int> dx = {1, 0, -1, 0}, dy = {0, 1, 0, -1};
//vector<int> dx2 = { 1,1,0,-1,-1,-1,0,1 }, dy2 = { 0,1,1,1,0,-1,-1,-1 };



/*--------additional--------*/



/*----------main------------*/

int main() {
/*----------fastio----------*/
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    std::ios::sync_with_stdio(false);
/*----------solve-----------*/

    int k, n; cin >> k >> n;
    vector<int> a(n);
    REP(i, n) cin >> a[i];

    int tmp = a[0] - a[n-1];

    for(int i = 0; i < n-1; i++){
        a[i+1] -= a[i];
    }
    a[0] = tmp;


    sort(rALL(a));
    k -= a[0];

    cout << k << endl;

return 0;
}