Problem:
Score : 400 points

Problem Statement

We have N integers. The i-th integer is A_i.

Find \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

What is \mbox{ XOR }?

The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:

When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.

For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)

Constraints

2 \leq N \leq 3 \times 10^5

0 \leq A_i < 2^{60}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the value \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

Sample Input 1

3
1 2 3

Sample Output 1

6

We have (1\mbox{ XOR } 2)+(1\mbox{ XOR } 3)+(2\mbox{ XOR } 3)=3+2+1=6.

Sample Input 2

10
3 1 4 1 5 9 2 6 5 3

Sample Output 2

237

Sample Input 3

10
3 14 159 2653 58979 323846 2643383 27950288 419716939 9375105820

Sample Output 3

103715602

Print the sum modulo (10^9+7).

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

#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<stdio.h>
#include<vector>
#include<queue>
#include<math.h>
using namespace std;
#define int long long
#define double long double
#define rep(s,i,n) for(int i=s;i<n;i++)
#define c(n) cout<<n<<endl;
#define ic(n) int n;cin>>n;
#define sc(s) string s;cin>>s;
#define mod 1000000007
#define inf 1000000000000000007
#define f first
#define s second
#define mini(c,a,b) *min_element(c+a,c+b)
#define maxi(c,a,b) *max_element(c+a,c+b)
#define pi 3.141592653589793238462643383279
#define e_ 2.718281828459045235360287471352
#define P pair<int,int>
#define upp(a,n,x) upper_bound(a,a+n,x)-a;
#define low(a,n,x) lower_bound(a,a+n,x)-a;
#define UF UnionFind 
//printf("%.12Lf\n",);
int keta(int x){
	rep(0,i,30){
		if(x<10){
			return i+1;
		}
		x=x/10;
	}
}	
int gcd(int x,int y){
	int aa=x,bb=y;
	rep(0,i,1000){
		aa=aa%bb;
		if(aa==0){
			return bb;
		}	
		bb=bb%aa;
		if(bb==0){
			return aa;
		}
	}
}	
int lcm(int x,int y){
	int aa=x,bb=y;
	rep(0,i,1000){
		aa=aa%bb;
		if(aa==0){
			return x/bb*y;
		}	
		bb=bb%aa;
		if(bb==0){
			return x/aa*y;
		}
	}
}
bool p(int x){
	if(x==1)return false;
	rep(2,i,sqrt(x)+1){
		if(x%i==0&&x!=i){
			return false;
		}
	}	
	return true;
}
int max(int a,int b){
	if(a>=b)return a;
	else return b;
}
string maxst(string s,string t){
	int n=s.size();
	int m=t.size();
	if(n>m)return s;
	else if(n<m)return t;
	else{
		rep(0,i,n){
			if(s[i]>t[i])return s;
			if(s[i]<t[i])return t;
		}
		return s;
	}
}		
int min(int a,int b){
	if(a>=b)return b;
	else return a;
}	
int n2[41];		
int nis[41]; 
int nia[41];
int mody[41];
int nn;
int com(int n,int y){
	int ni=1;
	for(int i=0;i<41;i++){
		n2[i]=ni;
		ni*=2;
	}
	int bunsi=1,bunbo=1;
	rep(0,i,y)bunsi=(bunsi*(n-i))%mod;
	rep(0,i,y)bunbo=(bunbo*(i+1))%mod;
	mody[0]=bunbo;
	rep(1,i,41){
		bunbo=(bunbo*bunbo)%mod;
		mody[i]=bunbo;
	}
	rep(0,i,41)nis[i]=0;
	nn=mod-2;
	for(int i=40;i>=0;i-=1){
		if(nn>n2[i]){
			nis[i]++;
			nn-=n2[i];
		}
	}
	nis[0]++;
	rep(0,i,41){
		if(nis[i]==1){
			bunsi=(bunsi*mody[i])%mod;
		}
	}
	return bunsi;
}
int gyakugen(int n,int y){
	int ni=1;
	for(int i=0;i<41;i++){
		n2[i]=ni;
		ni*=2;
	}
	mody[0]=y;
	rep(1,i,41){
		y=(y*y)%mod;
		mody[i]=y;
	}
	rep(0,i,41)nis[i]=0;
	nn=mod-2;
	for(int i=40;i>=0;i-=1){
		if(nn>n2[i]){
			nis[i]++;
			nn-=n2[i];
		}
	}
	nis[0]++;
	rep(0,i,41){
		if(nis[i]==1){
			n=(n*mody[i])%mod;
		}
	}
	return n;	
}	
int yakuwa(int n){
	int sum=0;
	rep(1,i,sqrt(n+1)){
		if(n%i==0)sum+=i+n/i;
		if(i*i==n)sum-=i;
	}
	return sum;
}	
bool integer(double n){
	if(n==long(n))return true;
	else return false;	
}
int poow(int y,int n){	
	if(n==0)return 1;
	n-=1;
	int ni=1;
	for(int i=0;i<41;i++){
		n2[i]=ni;
		ni*=2;
	}
	int yy=y;
	mody[0]=yy;
	rep(1,i,41){
		yy=(yy*yy)%mod;
		mody[i]=yy;
	}
	rep(0,i,41)nis[i]=0;
	nn=n;
	for(int i=40;i>=0;i-=1){
		if(nn>=n2[i]){
			nis[i]++;
			nn-=n2[i];
		}
	}
	rep(0,i,41){
		if(nis[i]==1){
			y=(y*mody[i])%mod;
		}
	}
	return y;	
}
int minpow(int x,int y){
	int sum=1;
	rep(0,i,y)sum*=x;
	return sum;
}
int ketawa(int x,int sinsuu){
	int sum=0;
	rep(0,i,100)sum+=(x%poow(sinsuu,i+1))/(poow(sinsuu,i));
	return sum;
}	
int sankaku(int a){
	return a*(a+1)/2;
} 
int same(int a[1111111],int n){
	int ans=0;
	rep(0,i,n){
		if(a[i]==a[i+1]){
			int j=i;
			while(a[j+1]==a[i]&&j<=n-2)j++;
			ans+=sankaku(j-i);
			i=j;
		}
	}
	return ans;
}
bool search(int x){
	int n;
	int a[n];
	int l=0,r=n;
	while(r-l>=1){
		int i=(l+r)/2;
		if(a[i]==x)return true;
		else if(a[i]<x)l=i+1;
		else r=i;
	}
	return false;
}
/*using Graph=vector<vector<int>>;
int oya[114514];
int depth[114514];
void dfs(const Graph &G, int v, int p, int d) {
    depth[v] = d;
    oya[v]=p;
    for (auto nv : G[v]) {
       if (nv == p) continue; // nv が親 p だったらダメ
       dfs(G, nv, v, d+1); // d を 1 増やして子ノードへ
    }
}*/
/*int H=10,W=10;
char field[10][10];
char memo[10][10];
void dfs(int h, int w) {
    memo[h][w] = 'x';

    // 八方向を探索
    for (int dh = -1; dh <= 1; ++dh) {
        for (int dw = -1; dw <= 1; ++dw) {
        	if(abs(0-dh)+abs(0-dw)==2)continue;
            int nh = h + dh, nw = w + dw;

            // 場外アウトしたり、0 だったりはスルー
            if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;
            if (memo[nh][nw] == 'x') continue;

            // 再帰的に探索
            dfs(nh, nw);
        }
    }
}*/
int XOR(int a,int b){
	int ni=1;
	rep(0,i,41){
		n2[i]=ni;
		ni*=2;
	}
	rep(0,i,41)nis[i]=0;
	for(int i=40;i>=0;i-=1){
		if(a>=n2[i]){
			nis[i]++;
			a-=n2[i];
		}
		if(b>=n2[i]){
			nis[i]++;
			b-=n2[i];
		}	
	}
	int sum=0;
	rep(0,i,41)sum+=(nis[i]%2*n2[i]);
	return sum;
}
//int ma[1024577][21];
//for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=1;
struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};
int a[314514];
int cnt[314514][60];
int sum[60];
int ans[60];
signed main(){
	ic(n) 		
	rep(0,i,n)cin>>a[i];
	rep(0,i,n)rep(0,j,60)cnt[i][j]=0;
	rep(0,i,n){
		int k=1;
		rep(0,j,60){
			if(a[i]%(k*2)/k==1){
				cnt[i][j]++;
				sum[j]++;
			}	
			k*=2;
		}
	}
	rep(0,i,60)ans[i]=(sum[i]*(n-sum[i]))%mod;
	int answer=0;
	int l=1;
	rep(0,i,60){
		answer+=l*ans[i];
		l*=2;
		answer%=mod;
	}
	c(answer)
}