#include<bits/stdc++.h>

#define int long long
using namespace std;
int arr[2000009];

inline int exgcd(int a, int b, int &x, int &y)
{
	if(!b){x = 1;y = 0;}
	else
	{
		exgcd(b, a % b, x, y);
		int t = x;
		x = y;y = t - (a / b) * y;
	}
	return 0;
}

void solve()
{

    int n,m;
    scanf("%lld%lld",&n,&m);
    int ans,s,d;
    for(int i=1;i<=n;i++)
    {
       scanf("%lld",&arr[i]);
    }
    int sum=0;
    for(int i=1;i<=n;i++)
    {
       sum+=arr[i];
       sum%=m;
    }
    int tmpsum=sum,tmpa=n%m,tmpb=(n+1)*n/2;
    tmpb%=m;
    if(sum==0)
    {
        ///printf("0\n");
        ans=0;
        s=0,d=0;
        ///printf("0 0\n");
    }
    else
    {
        int a=n%m,b=(n+1)*n/2;
        b%=m;
        if(a==0&&b==0)
        {
            ans=sum;
            s=0,d=0;
            //printf("%lld\n",sum);
            //printf("0 0\n");
        }
        else if(a==0)
        {
           int k=(m-sum+b-1)/b;
           int ans1=sum,ans2=(sum+k*b)%m;
           if(ans1<=ans2)
           {
               ans=ans1;s=0,d=0;
               ///printf("%lld\n",ans1);
               ///printf("0 0\n");
           }
           else
           {
               ans=ans2;
               s=0;d=k;
              ///printf("%lld\n",asn2);
              ///printf("0 %lld\n",k);
           }
        }
        else if(b==0)
        {
           swap(a,b);
            int k=(m-sum+b-1)/b;
           int ans1=sum,ans2=(sum+k*b)%m;
           if(ans1<=ans2)
           {
               ans=ans1;
               s=0,d=0;
              /// printf("%lld\n",ans1);
               ///printf("0 0\n");
           }
           else
           {
               ans=ans2;
               s=k,d=0;
              ///printf("%lld\n",asn2);
              ///printf("%lld 0\n",k);
           }
        }
        else
        {
            sum=m-sum;///这是需求;
           int gd=__gcd(a,b);
           if((sum%gd)==0)
           {
               a/=gd,b/=gd,sum/=gd;
              int x,y;///a*x+b*y=c;
              exgcd(a,b,x,y);
              x%=m;
              x+=m;
              x%=m;
              y%=m;
              y+=m;
              y%=m;
              x*=sum;x%=m;
              y*=sum;y%=m;
              ans=0;
              s=x,d=y;
        //  printf("ab:%lld %lld %lld %lld\n",a,b,x,y);
           }
           else
           {
              sum=m-sum;
              int k=(m-sum+gd-1)/gd;
              if(sum<=(sum+k*gd%m)%m)
              {
                  ans=sum;
                  s=0,d=0;
              }
              else
              {
                 int x,y;
                 exgcd(a,b,x,y);
                  x%=m;
                  x+=m;
                  x%=m;
                 y%=m;
                 y+=m;
                 y%=m;
                x*=k;x%=m;
                y*=k;y%=m;
                ans=(sum+k*gd)%m;
                s=x,d=y;

              }
           }
        }
    }
    ///这边是check(),函数的意义;
    int ss=0;
    for(int i=1;i<=n;i++)
    {
       ss+=s+arr[i]+i*d%m;
       ss%=m;
    }
    if(ss!=ans)
    {
        printf("%lld\n",arr[30400500405]);
      // printf("WA ss,%lld ans,%lld\n",ss,ans);
      // printf("tmpsum,%lld tmpa:%lld tmpb:%lld s:%lld d:%lld\n",tmpsum,tmpa,tmpb,s,d);
    }
    else
    {
        printf("%lld\n",ans);
        printf("%lld %lld",s,d);
    }
}


signed main()
{
     int tt=1;
    // scanf("%lld",&tt);
     while(tt--)
     {
       solve();
     }
    return 0;
}