#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pii pair<int, int>
const int N = 1e6;
const int M = 2e3;
int pow(int a, int b, int m)
{
    int ans = 1;
    a %= m;
    while (b)
    {
        if (b & 1)
            ans = (ans % m) * (a % m) % m;
        b /= 2;
        a = (a % m) * (a % m) % m;
    }
    ans %= m;
    return ans;
}
int extgcd(int a, int b, int &x, int &y)
// 求解ax+by=gcd(a, b)
// 返回值为gcd(a, b)
{
    int d = a;
    if (b)
    {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    }
    else
        x = 1, y = 0;
    return d;
}
int mod_inverse(int a, int m)
// 求解a关于模上m的逆元
// 返回-1表示逆元不存在
{
    int x, y;
    int d = extgcd(a, m, x, y);
    return d == 1 ? (m + x % m) % m : -1;
}
void extend_gcd(int a, int b, int &d, int &x, int &y)
{
    if (!b)
    {
        d = a;
        x = 1;
        y = 0;
    }
    else
    {
        extend_gcd(b, a % b, d, y, x);
        y -= x * (a / b);
    }
}
void solve()
{
    int n, m;
    cin >> n>>m;
    vector<int> a(n+1);
    int sum = 0;
    for (int i = 1; i <= n;i++){
        cin >> a[i];
        sum += a[i];
    }
    if(sum%m==0){
        cout << 0 << endl
             << 0 << " " << 0 << endl;
        return;
    }
    int g = __gcd(n, n * (n + 1) / 2);
    int k = (m - sum  + g - 1) / g;
    int s, d;
    int t;
    extend_gcd(n, n * (n + 1) / 2, t, s, d);
    int b1 = n * (n + 1) / 2 / t;
    int x1 = (s + b1) * (k*g / d);
    x1 = (x1 % b1 + b1) % b1;
    int y1 = (g * k - n * x1) / n * (n + 1) / 2;
    cout
        << (sum + k * g) % m << endl;
    cout << x1 << " " << y1 << endl;
    return;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t = 1;
    //cin >> t;
    while (t--)
        solve();
    return 0;
}