Problem:
Score : 300 points

Problem Statement

Given are an integer X and an integer sequence of length N: p_1, \ldots, p_N.

Among the integers not contained in the sequence p_1, \ldots, p_N (not necessarily positive), find the integer nearest to X, that is, find the integer whose absolute difference with X is the minimum. If there are multiple such integers, report the smallest such integer.

Constraints

1 \leq X \leq 100

0 \leq N \leq 100

1 \leq p_i \leq 100

p_1, \ldots, p_N are all distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X N
p_1 ... p_N

Output

Print the answer.

Sample Input 1

6 5
4 7 10 6 5

Sample Output 1

8

Among the integers not contained in the sequence 4, 7, 10, 6, 5, the one nearest to 6 is 8.

Sample Input 2

10 5
4 7 10 6 5

Sample Output 2

9

Among the integers not contained in the sequence 4, 7, 10, 6, 5, the ones nearest to 10 are 9 and 11. We should print the smaller one, 9.

Sample Input 3

100 0

Sample Output 3

100

When N = 0, the second line in the input will be empty. Also, as seen here, X itself can be the answer.

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

typedef int keytype;

void inssort(int n, keytype a[])
{
    int i, j;
    keytype x;

    for (i = 1; i < n; i++) {
        x = a[i];
        for (j = i - 1; j >= 0 && a[j] > x; j--)
            a[j + 1] = a[j];
        a[j + 1] = x;
    }
}

#define THRESHOLD 10
#define STACKSIZE 32  /* たかだか int のビット数程度 */

void quicksort(int n, keytype a[])
{
    int i, j, left, right, p;
    int leftstack[STACKSIZE], rightstack[STACKSIZE];
    keytype x, t;

    left = 0;  right = n - 1;  p = 0;
    for ( ; ; ) {
        if (right - left <= THRESHOLD) {
            if (p == 0) break;
            p--;
            left = leftstack[p];
            right = rightstack[p];
        }
        x = a[(left + right) / 2];
        i = left;  j = right;
        for ( ; ; ) {
            while (a[i] < x) i++;
            while (x < a[j]) j--;
            if (i >= j) break;
            t = a[i];  a[i] = a[j];  a[j] = t;
            i++;  j--;
        }
        if (i - left > right - j) {
            if (i - left > THRESHOLD) {
                leftstack[p] = left;
                rightstack[p] = i - 1;
                p++;
            }
            left = j + 1;
        } else {
            if (right - j > THRESHOLD) {
                leftstack[p] = j + 1;
                rightstack[p] = right;
                p++;
            }
            right = i - 1;
        }
    }
    inssort(n, a);
}

int main(){
    int x, n;
    int p[110];

    scanf("%d %d", &x, &n);
    for(int i = 0; i < n; i++){
        scanf("%d", &p[i]);
    }

    quicksort(n, p);

    /*
    for(int i = 0; i < n; i++){
        printf("%d ", p[i]);
    }
    puts("");
    */
    int flag = 0, ans = x;
    int i;
    for(i = 0; i < n; ++i){
        if(p[i] == x){
            flag = 1;
            break;
        }
    }
    printf("%d\n", i);
    if(flag){
        int j = 1;
        while((i - j) > 0 || (i + j) < n){
            if((i - j) > 0 && (x - j) != p[i - j]){
                ans = x - j;
                break;
            }
            if((i + j) < n && (x + j) != p[i + j]){
                ans = x + j;
                break;
            }
            j++;
        }
    }

    printf("%d\n", ans);

    return 0;
}