#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<string>
#include<cstring>
#include<set>
#include<cmath>
#include<queue>
#define int long long
#define endl '\n'
#define F first
#define S second
#define sz size()
#define B begin()
#define D end()
#define pii pair<int,int>
#define IOS ios::sync_with_stdio(false)

using namespace std;
const int N=1e6+10;
const int mod=1e9+7;

int ksm(int m,int k,int mod);
int gcd(int a,int b);
int exgcd(int a,int b,int &x,int &y);

int n,m;
int a[N];

void solve()
{
	cin>>n>>m;
	int sum=0;
	for(int i=1;i<=n;i++){
		int a;
		cin>>a;
		sum+=a;
	}
	int k=gcd(n,n*(n+1)/2);
	int g=gcd(k,m);
	int f=(-sum)/g;
	int ans_f=f*g;
	int ans=ans_f+sum;
	int x,y;
	exgcd(k,m,x,y);
	x*=f;
	y*=f;
	int s,d;
	exgcd(n,(n+1)*n/2,s,d);
	s=s/g*(ans_f-m*y);
	d=d/g*(ans_f-m*y);
	cout<<ans<<endl;
	cout<<s<<' '<<d<<endl;
}
signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int tc=1;
	//cin>>tc;
	while(tc--)
	{
		solve();
	}
	return 0;
}

int ksm(int m,int k,int mod)
{
	int ans=1;
	int t=m;
	while(k)
	{
		if(k&1) ans=ans*t%mod;
		t=t*t%mod;
		k>>=1;
	}
	return ans;
}

int gcd(int a,int b)
{
	return b==0?a:gcd(b,a%b);
}

int exgcd(int a,int b,int& x,int& y)
{
	if(b==0){
		x=1,y=0;
		return a;//返回值为gcd（a，b）
	}
	int res=exgcd(b,a%b,x,y);//迭代到最底层
	int t=x;
	x=y;
	y=t-(a/b)*y;
	//完成实现时的x1=y2,y1=x2-(a/b)*y2
	//由于是从最深层实现的，故直接用公式向上层传递即可
	return res;
}