Problem:
Score : 600 points

Problem Statement

We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller.

Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.

It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations.

You are given an integer K. Find an integer sequence a_i such that the number of times we will perform the above operation is exactly K. It can be shown that there is always such a sequence under the constraints on input and output in this problem.

Constraints

0 ≤ K ≤ 50 \times 10^{16}

Input

Input is given from Standard Input in the following format:

K

Output

Print a solution in the following format:

N
a_1 a_2 ... a_N

Here, 2 ≤ N ≤ 50 and 0 ≤ a_i ≤ 10^{16} + 1000 must hold.

Sample Input 1

0

Sample Output 1

4
3 3 3 3

Sample Input 2

1

Sample Output 2

3
1 0 3

Sample Input 3

2

Sample Output 3

2
2 2

The operation will be performed twice: [2, 2] -> [0, 3] -> [1, 1].

Sample Input 4

3

Sample Output 4

7
27 0 0 0 0 0 0

Sample Input 5

1234567894848

Sample Output 5

10
1000 193 256 777 0 1 1192 1234567891011 48 425

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,s,n) for(int i=s;i<n;i++)
#define pb push_back
#define all(x) (x).begin(),(x).end()
using namespace std;
using P=pair<int,int>;
using Graph=vector<vector<int>>;
using ll=long long;
ll gcd(ll A, ll B){
    if(B == 0)return A;
    return gcd(B, A % B);
}
ll lcm(ll A, ll B){
    return A * B / gcd(A, B);
}
bool root(string s){
    int n=s.length();
    if(n==2){
        if(s[0]==s[1]){
            return true;
        }
        else return false;
    }
    if(n==3){
        if(s[0]==s[2]){
            return true;
        }
        else return false;
    }

    return false;
}


//深さ優先
vector<int> seen;
vector<ll> h;

//普通の深さ優先探索seenを返す感じ
void dfs(const Graph &G,int v){
    seen[v] = 0; //v探索済み

    P good=make_pair(v,h[v]);
    for(auto next_v :G[v]){
        if(seen[next_v]!=-1) continue;//もう探索していた
        if(h[next_v]>good.second){
            good=make_pair(next_v,h[next_v]);
        }
        dfs(G,next_v);
    }

    if(!(h[v]==good.second && v==good.first)){
        seen[good.first]=1;
    }
}


int main(){
    /*
    int n,m;
    cin>>n>>m;

    h.assign(n,0);
    rep(i,0,n) cin>>h[i];

    Graph G(n);
    rep(i,0,m){
        int a,b; cin>>a>>b;
        G[a-1].push_back(b-1);
        G[b-1].push_back(a-1);
    }

    seen.assign(n,-1);

    rep(i,0,n){
        if(seen[i]==-1){
            dfs(G,i);
        }
    }

    int sum=accumulate(all(seen),0);

    cout<<sum<<endl;*/

    ll k; cin>>k;
    ll n=50;
    ll s[n],ans[n];
    vector<ll> d(n,0);

    ll div=k/n; ll ama=k%n;

    rep(i,0,n){
        s[i]=div+(i<ama);
    }

    rep(i,0,n){
        rep(j,0,n){
            if(i!=j);
                d[i]+=s[j];
        }
    }

    rep(i,0,n){
        ans[i]=max(0LL,n*s[i]-d[i]);
    }

    rep(i,0,n){
        cout<<(i ? " ":"") <<ans[i];
    }

    cout<<endl;
}