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 _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define dump(x) cout << x << endl
#define answer(x) cout << x << endl
typedef int64_t Int;

using namespace std;
using Graph = vector<vector<Int>>;

const double pi = M_PI;
const Int MOD = 1000000007;

Int modpow(Int a, Int n, Int mod) {
    Int ret = 1;
    while (n > 0) {
        if (n & 1) {
            ret = ret * a % mod;
        }
        a = a * a % mod;
        n >>= 1;
    }

    return ret;
}

vector<Int> factorials(0), invs(0);
bool is_ncr_table = false;
void calc_ncr_table(Int n, Int mod) {
    is_ncr_table = true;

    // 1! ~ n! の計算
    Int f = 1;
    factorials.push_back(f);
    for (Int i = 1; i < n + 1; i++) {
        f *= i;
        f %= mod;
        factorials.push_back(f);
    }
    // n!^-1 の計算
    Int inv = modpow(factorials[n], mod - 2, mod);
    // (n-1)!^-1 ~ 1!^-1 の計算
    invs.assign(n, 1);
    invs[n] = inv;
    for (Int i = n; i >= 1; i--) {
        inv *= i;
        inv %= mod;
        invs[i - 1] = inv;
    }
}

Int factorial(Int n, Int mod = 1) {
    Int ret = 1;
    if (mod == 1) {
        for (Int i = 1; i < n + 1; i++) {
            ret *= i;
        }

        return ret;
    } else {
        for (Int i = 1; i < n + 1; i++) {
            ret *= i;
            ret %= mod;
        }

        return ret;
    }
}

Int ncr(Int n, Int r, Int mod, bool use_table = false) {
    if (use_table) {
        // modが素数の場合のみ
        if (!is_ncr_table) {
            calc_ncr_table(n, mod);
        }
        return factorials[n] * invs[r] * invs[n - r] % mod;
    } else {
        Int x = 1;
        for (Int i = n - r + 1; i < n + 1; i++) {
            x *= i;
            x %= MOD;
        }
        return x * modpow(factorial(r, mod), mod - 2, mod) % mod;
    }
}

int main() {
    int K, N;
    cin >> K >> N;
    vector<int> A(N);
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }

    vector<int> dif(N);
    for (int i = 1; i < N; i++) {
        dif[i] = A[i] - A[i - 1];
    }
    dif[0] = K - A[N - 1] + A[0];
    sort(dif.begin(), dif.end());

    cout << K - dif[N - 1] << endl;

    return 0;
}