Problem:
Score : 200 points

Problem Statement

There are K pieces of cakes.
Mr. Takahashi would like to eat one cake per day, taking K days to eat them all.

There are T types of cake, and the number of the cakes of type i (1 ≤ i ≤ T) is a_i.

Eating the same type of cake two days in a row would be no fun,
so Mr. Takahashi would like to decide the order for eating cakes that minimizes the number of days on which he has to eat the same type of cake as the day before.

Compute the minimum number of days on which the same type of cake as the previous day will be eaten.

Constraints

1 ≤ K ≤ 10000

1 ≤ T ≤ 100

1 ≤ a_i ≤ 100

a_1 + a_2 + ... + a_T = K

Input

The input is given from Standard Input in the following format:

K T
a_1 a_2 ... a_T

Output

Print the minimum number of days on which the same type of cake as the previous day will be eaten.

Sample Input 1

7 3
3 2 2

Sample Output 1

0

For example, if Mr. Takahashi eats cakes in the order of 2, 1, 2, 3, 1, 3, 1, he can avoid eating the same type of cake as the previous day.

Sample Input 2

6 3
1 4 1

Sample Output 2

1

There are 6 cakes.
For example, if Mr. Takahashi eats cakes in the order of 2, 3, 2, 2, 1, 2, he has to eat the same type of cake (i.e., type 2) as the previous day only on the fourth day.
Since this is the minimum number, the answer is 1.

Sample Input 3

100 1
100

Sample Output 3

99

Since Mr. Takahashi has only one type of cake, he has no choice but to eat the same type of cake as the previous day from the second day and after.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define ld long double
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
using P=pair<long long,long long>;
#define rep(i,n) for(long long i=0; i<(long long)n; i++)
#define req(i,n) for(long long i=n-1; i>=0; i--)
#define range(i,a,b) for(long long i=a; i<b; i++)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((long long)(x).size())
#define COUT(x) cout << x << endl
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define onBoard(y,x) (y>=0 && y<h && x>=0 && x<w)
#define vint vector<int>
#define pri_que priority_queue
#define vvint vector<vector<int>>
#define vstr vector<string>
#define vp vector<pair<int,int>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define show(x) cout<<#x<<"="<<x<<endl;
#define SUM(x) accumulate(x.begin(),x.end(),0)
#define MAX(x) *max_element(x.begin(),x.end())
#define MIN(x) *min_element(x.begin(),x.end())
#define couty cout<<"Yes"<<endl
#define coutn cout<<"No"<<endl
#define coutY cout<<"YES"<<endl
#define coutN cout<<"NO"<<endl
#define yn(x) cout<<(x?"Yes":"No")<<endl
#define YN(x) cout<<(x?"YES":"NO")<<endl
long long gcd(long long a,long long b){return b?gcd(b,a%b):a;}
long long lcm(long long a, long long b){return a/gcd(a,b)*b;}
const long long dx[8]={1,0,-1,0,1,-1,-1,1};
const long long dy[8]={0,1,0,-1,1,1,-1,-1};
const long long INF = 1e12;
const long long MOD = 1e9+7;

signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout<<fixed<<setprecision(15);

    int n,m; cin >> m >> n;
    vint a(n); rep(i,n) cin >> a[i];

    int ma=MAX(a),other=SUM(a)-ma;
    cout<<max(0LL,ma-other-1)<<endl;   
}