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)

//Mahir Ratanpara (DA-IICT)
#include<bits/stdc++.h>
using namespace std;
#define Ff(i,a,n) for(i=a;i<n;i++)
#define Fr(i,a,n) for(i=a;i>n;i--)
#define ll long long
#define FAST ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
typedef pair<ll, ll>	pa;
typedef vector<ll>		vc;
typedef vector<pa>		vp;
typedef vector<vc>		vvc;
int mpow(int base, int exp); 
void ipgraph(ll n,ll m);
void dfs(int u, int par);
const int mod = 1000000007;
const int N = 3e5, M = N;
//=======================

vc g[N];

int main() {
    FAST;
    ll t, c,i, j, k, p, q, r, x, y, u, v, n, m;
	cin>>x>>n;
	vc mp(101,0);
	Ff(i,0,n)
	{
		cin>>y;
		mp[y]++;
	}

	y=x;
	while(y<=101)
	{
		if(mp[y]>0)
		y++;
		else{
			break;
		}
	}

	u=x;
	while(u>=0)
	{
		if(mp[u]>0)
		u--;
		else{
			break;
		}
	}

	if(x-u<=y-x)
	cout<<u;
	else
	cout<<y;

    return 0;
}

int mpow(int base, int exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((ll)result * base) % mod;
    base = ((ll)base * base) % mod;
    exp >>= 1;
  }
  return result;
}

void ipgraph(ll n, ll m){
	int i, u, v;
	while(m--){
		cin>>u>>v;
		g[u-1].pb(v-1);
		g[v-1].pb(u-1);
	}
}

void dfs(int u, int par){
	for(int v:g[u]){
		if (v == par) continue;
		dfs(v, u);
	}
}