#include<bits/stdc++.h>
using namespace std;
typedef int64_t LL;
#define rep(i,N) for(LL i=0;i<(int)N;i++)
constexpr int MOD=1e9+7;
constexpr LL SCORE=1LL<<20,Z=(1LL<<0)|(1LL<<20)|(1LL<<40);
int main()
{
  int N;
  LL A=0,x,y;
  vector<LL> bit(20,0);
  cin>>N;
  while(cin>>x)rep(i,20)bit.at(i)+=x>>i&Z,x>>1;
  rep(i,20){x=bit.at(i);rep(j,3){y=x&SCORE-1;x/=SCORE;A=fmod(A+fmod(y%MOD*(N-y)%MOD*pow(2,i+20*j),MOD),MOD);}}
  return cout<<A<<endl,0;
}
### Prompt

Your challenge is to write a CPP solution to the following problem:

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).

Examples

Input

3
1 2 3


Output

6


Input

10
3 1 4 1 5 9 2 6 5 3


Output

237


Input

10
3 14 159 2653 58979 323846 2643383 27950288 419716939 9375105820


Output

103715602

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef int64_t LL;
#define rep(i,N) for(LL i=0;i<(int)N;i++)
constexpr int MOD=1e9+7;
constexpr LL SCORE=1LL<<20,Z=(1LL<<0)|(1LL<<20)|(1LL<<40);
int main()
{
  int N;
  LL A=0,x,y;
  vector<LL> bit(20,0);
  cin>>N;
  while(cin>>x)rep(i,20)bit.at(i)+=x>>i&Z,x>>1;
  rep(i,20){x=bit.at(i);rep(j,3){y=x&SCORE-1;x/=SCORE;A=fmod(A+fmod(y%MOD*(N-y)%MOD*pow(2,i+20*j),MOD),MOD);}}
  return cout<<A<<endl,0;
}
```