#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef double db;
#define int long long
const int N = 2e5 + 5;
const double eps = 1e-10;
int x, y;
//ax+by=c
//裴蜀定理得c%gcd(a,b)==0才有解 解ax=c（mod b）=ax+by=c 可解得x=x0*c/gcd（a,b）,此时mod（b(b值会变化)/gcd（a,b)）
int exgcd(int a, int b, int &x, int &y) {
    if (b == 0) {
        x = 1, y = 0;
        return a;
    }
    int x1, y1;
    int d = exgcd(b, a % b, x1, y1);
    x = y1, y = x1 - a / b * y1;
    return d;
}

void solve() {
    int n, m;
    cin >> n >> m;
    int s = 0;
    for (int i = 1, x; i <= n; i++) {
        cin >> x;
        s += x;
        s %= m;
    }
    int a = n;
    x = 0, y = 0;
    int d = exgcd(a, m, x, y);
    s = m - s;
    if (s % d == 0) {
        x*=s/d;
        cout << 0 << '\n';
        int l = m / d;
        x %= l;
        x += l;
        x %= l;
        cout << x << " " << 0 << '\n';
        return;
    }
    a = (n) * (n + 1) / 2;
    x = 0, y = 0;
    d = exgcd(a, m, x, y);
    if (s % d == 0) {
        x*=s/d;
        cout << 0 << '\n';
        int l = m / d;
        x %= l;
        x += l;
        x %= l;
        cout << 0 << " " << x << '\n';
        return;
    }
    for(int i=1;i<=n;i++){
        s++;
        s %= m;
        a = n;
        x = 0, y = 0;
        d = exgcd(a, m, x, y);
        if (s % d == 0) {
            x*=s/d;
            cout << i << '\n';
            int l = m / d;
            x %= l;
            x += l;
            x %= l;
            cout << x << " " << 0 << '\n';
            return;
        }
        a = (n) * (n + 1) / 2;
        x = 0, y = 0;
        d = exgcd(a, m, x, y);
        if (s % d == 0) {
            x*=s/d;
            cout << i << '\n';
            int l = m / d;
            // cout<<x<<" "<<l<<'\n';
            x %= l;
            x += l;
            x %= l;
            cout << 0 << " " << x << '\n';
            return;
        }
    }
}


signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    //  cout << fixed << setprecision(10);
    int T = 1;
    //cin >> T;
    while (T--) solve();
    return 0;
}