#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl "\n"
#define ull unsigned long long
#define ms(x, y) memset(x, y, sizeof x);
#define debug(x) cout << #x << " = " << x << endl;
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#define fre                           \
freopen("input.txt", "r", stdin); \
freopen("output.txt", "w", stdout);
const int mod = 998244353;
const int inf = 0x3f3f3f3f3f3f3f3f;
const int N = 1e6 + 10;
const double esp = 1e-6;
const ull MOD1 = 1610612741;
const ull MOD2 = 805306457;
const ull BASE1 = 1331;
const ull BASE2 = 131;
#define pre(i, a, b) for (int i = a; i <= b; i++)
#define rep(i, a, b) for (int i = a; i >= b; i--)
#define all(x) (x).begin(), (x).end()
char *p1, *p2, buf[100000]; // 快读和同步流二者只能选一个
#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++)
int read()
{
	int x = 0, f = 1;
	char ch = nc();
	while (ch < 48 || ch > 57)
	{
		if (ch == '-')
			f = -1;
		ch = nc();
	}
	while (ch >= 48 && ch <= 57)
		x = x * 10 + ch - 48, ch = nc();
	return x * f;
}
void write(int x)
{
	if (x < 0)
		putchar('-'), x = -x;
	if (x > 9)
		write(x / 10);
	putchar(x % 10 + '0');
	return;
}
int n,m;
int a[N];
int exgcd(int a,int b,int &x,int &y)//扩展欧几里得算法 ,求出来的结果是方程变换后右值等于 gcd 的系数 x y
{
	if(b == 0)
	{
		x = 1,y = 0;
		return a;
	}
	int ret = exgcd(b,a%b,y,x);
	y -= a / b * x;
	return ret;
}
void solve()
{
	cin >> n >> m;
    int sum = 0;
    pre(i,1,n) cin >> a[i],sum += a[i];
    sum %= m;
    int d1 = __gcd(n * (n + 1) / 2,n);
    int d2 = n * (n + 1) / 2,d3 = n;
    if(sum == 0)
    {
        cout << 0 << "\n" << "0 0" << endl;
        return ;
    }
    sum = m - sum;
    if(sum % d1 == 0)
    {
        cout << 0 << endl << 1 << endl << 1 << endl;
        return ;
    }
    int ans = inf;
    pre(i,0,2 * d1)
    {
        if((sum + i) % d1 == 0)
        {
            ans = min(ans,(sum + i) % sum);
            // cout << i << endl;
            // return ;
        }
    }
        

    int x = 0,y = 0;
    //现在 的数据是 sum + ans
    exgcd(n * (n + 1) / 2,n,x,y);
    //求出来的结果是 右 = d 时的结果
    int tpa = (n * (n + 1) / 2) / d1,tpb = n / d1;
    x = x + (sum + ans + m) / d1 * tpa,y= y + (sum + ans + m) / d1 * tpb;

    
   
    x = (x + m) % m;
    y = (y + m) % m ;
    cout << ans << "\n" << x << " " << y << endl;


}
// #define LOCAL
signed main()
{
	ios
	// fre
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	auto start = std::chrono::high_resolution_clock::now();
#endif
	
	int t = 1;
	// cin >> t;
	while (t--)
		solve();
	
#ifdef LOCAL
	auto end = std::chrono::high_resolution_clock::now();
	cout << "Execution time: "
	<< std::chrono::duration_cast<std::chrono::milliseconds>(end - start).count()
	<< " ms" << '\n';
#endif
	return 0;
}