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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i=0;i<(n);++i)
#define reps(i,n) for(int i=1;i<=(n);++i)
#define repr(i,n) for(int i=(n);i>=0;--i)
#define step(i,s,n,c) for(int i=(s);i<(n);i+=(c))
#define REP(i,s,n) for(int i=(s);i<(n);++i)
#define each(i,vec) for(auto i:vec)
#define cinv(n,vec) VL vec(n);for(auto&v:vec)cin>>v;
#define int long long
#define itn int
#define pb push_back
#define eb emplace_back
#define m_p make_pair
#define m_t make_tuple
#define p_p(a,b) pb(m_p(a,b))
#define all(a) a.begin(),a.end()
#define SORT(a) sort(all(a))
#define RSORT(a) sort(all(a),greater<>())
#define UNIQUE(a) a.erase(unique(all(a)),a.end())
#define SZ(a) (signed)(a.size())
#define debg(a) cout<<#a<<" "<<a<<endl;
#define debgp(a) cout<<#a<<" "<<a.fi<<" : "<<a.se<<endl
#define call(a) for(auto i:a)cout<<i<<" ";cout<<endl
#define callp(a) for(auto [k,v]:a)cout<<k<<":"<<v<<endl
#define show(a) for(cont &y:a){for(cont &x:y){cout<<x<<" ";}cout<<endl;}
#define out(a) cout<<(a)<<endl
#define ENDL cout<<endl
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl
#define fi first
#define se second
#define V vector
#define P pair
#define T tuple
#define PQ priority_queue
#define cont const auto
#define VV(type,name,y,x,a) vector<vector<type>> name = vector<vector<type>>(y,vector<type>(x,a))
typedef long long ll;
typedef double D;
typedef long double LD;
typedef string str;
typedef vector<ll> VL;
typedef pair<ll,ll> PL;
typedef vector<pair<ll,ll>> VP;
typedef tuple<ll,ll,ll> T3;
typedef tuple<ll,ll,ll,ll> T4;
const signed INF = (1LL<<30);
const long long MOD = (1e9+7);
const long long LINF = (1LL<<60);
const long long LMAX = LLONG_MAX;
void YN(bool b){cout<<(b?"Yes":"No")<<endl;}
template<class T>inline T gcd(T a,T b){if(b==0)return a; return(gcd(b,a%b));}
template<class T>inline T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T>inline bool chmax(T &a,const T &b){if(a<b){a=b;return true;}return false;}
template<class T>inline bool chmin(T &a,const T &b){if(a>b){a=b;return true;}return false;}
ll vmax(VL v){int m;for(auto e:v)chmax(m,e);return m;}
signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(10);
  
  int k,t;
  cin>>k>>t;
  cinv(t,a);

  int mx=vmax(a);
  cout<<max(mx-1-(k-mx),0LL)<<endl;
  
  
  return 0;
}