#define _CRT_SECURE_NO_WARNINGS 1
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define double long double
//#define int long long
#define lc u<<1
#define rc u<<1|1
#define endl "\n"
#define X first
#define Y second
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> PII;
const ll N = 1e5 + 5;
const ll maxm = 2e18;
//const ll mod = 1e9 + 7;


ll exgcd(ll a, ll b, ll& x, ll& y) {
	if (b == 0) {
		x = 1; y = 0;
		return a;
	}
	ll g = exgcd(b, a % b, x, y);
	ll temp = x;
	x = y;
	y = temp - (a / b) * y;
	return g;
}
ll gcd(ll a, ll b) {
	return b ? gcd(b, a % b) : a;
}

//ll A[N];

void solve() {
	/*
    ll n,mod; cin >> n >> mod;
	//ll a, b, k, x, y; scanf("%lld%lld%lld", &a, &b, &k);
	ll a = n % mod, b = n * (n + 1) / 2 % mod;
	ll sum = 0;
	for (int i = 1; i <= n; i++)cin >> A[i], sum += A[i];
	ll x, y;
	ll d = exgcd(a, b, x, y);
	x = x * 1 / d;
	y = (1 - a * x) / b;
	cout << gcd(a, b) << endl;*/
	//刚才推的有问题，ans应该为:sum+k*gcd(gcd(a,b),m);,a=n,b=n*(n+1)/2;
	ll n, mod; cin >> n >> mod; ll sum = 0;
	for (int i = 1; i <= n; i++) {
		ll val; cin >> val, sum += val; sum %= mod;
	}
	ll a = n%mod, b = n * (n + 1) / 2%mod;
	/*
	for (int i = 1; i <= 10; i++) {
		cout << (sum + i * gcd(gcd(a, b), mod)) % mod << endl;
	}*/
	//原式为:k2*gcd(a,b)+k1*m=sum+k3*gcd(gcd(a,b),m)
	//=k4gcd(gcd(a,b),m),因为结果为任意一个，所以不妨直接取
	//感觉还是有问题，再推一个来
	//推到的式子是ans=k3*(gcd(gcd(a,b),m))+sum
	//可以看做是1*ans-k3*(gcd(gcd(a,b),m))=sum
	//不对，这样求出的是任意的ans，不是最小的ans
	///现在的问题是是要找到(sum+x*(gcd(gcd(a,b),m)))%mod的最小值
	//应该能出了
	//都快出了，思路又偏，ans=k3*(gcd(gcd(a,b),m))+sum
	//最小的ans应该为sum%(gcd(gcd(a,b),m))
	ll base = gcd(gcd(a, b), mod)%mod;
	ll ans = sum % mod; ans %= base;
	/*
	if (base % mod == 0||ans==0){}
	else {
		//ll dis = mod - base;
		//ll cnt = dis / base;
		//if (dis % base != 0)cnt++;
		//ll ex = cnt * base - mod;//这是在ans上一直放base
		ll k = (ans + mod - 1) / mod; // 向上取整
		ll min_value = (ans + k * mod - ans) % mod;
		ans = min_value;
	}*/
	//cout << ans << " " << base << endl;
	cout << ans << endl;
	//然后剩下就简单了
	//最后就是a*x+b*y+sum=ans-k1*mod
	//只要满足ans-sum-k1*mod=k2*gcd(a,b)即可
	//所以，ans-sum=k1*mod+k2*gcd(a,b)
	//好像也没有很简单，利用ans-sum=k1*mod+k2*gcd(a,b)解出k2
	//后面写的不好，
	ll x, y; exgcd(a, b, x, y);//然后看ans-sum-k1*mod是gcd(a,b)的几倍
	x %= mod; y %= mod;
	//cout << x << " " << y << endl;
	//ans已知，sum已知，此时ans-sum-k1*mod=k2*gcd(a,b)
	//也就是视为，k1*mod+k2*gcd(a,b)=ans-sum;
	//先算出k2,然后看看ans-sum是gcd(mod,gcd(a,b))的多少倍，
	ll k1, k2; exgcd(mod, gcd(a, b), k1, k2);
	//cout << k1 << " " << k2 << endl;
	k2 *= (ans - sum) / gcd(mod, gcd(a, b)); k2 %= mod;
	//cout << k2 << endl;
	/*
	x *= k2; x %= mod;
	y *= k2; y %= mod;*/
	x = (x * k2 % mod + mod) % mod;
	y = (y * k2 % mod + mod) % mod;
	cout << x << " " << y << endl;
    return;
}

signed main() {
    IOS;
    ll t = 1;
    //cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

/*
ll ksm(ll a, ll b) {
    ll base = a;
    ll ans = 1;
    while (b) {
        if (b & 1)ans *= base;
        base *= base;
        b >>= 1;
    }
    return ans;
}*/