codeforces 449D DP+容斥

Jzzhu and Numbers

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Appoint description: 
System Crawler  (2014-07-20)

Description

Jzzhu have n non-negative integers a₁, a₂, ..., a_n. We will call a sequence of indexes i₁, i₂, ..., i_k(1 ≤ i₁ < i₂ < ... < i_k ≤ n) a group of size k.

Jzzhu wonders, how many groups exists such that a_i1 & a_i2 & ... & a_ik = 0(1 ≤ k ≤ n)? Help him and print this number modulo1000000007(10⁹ + 7). Operation x & y denotes bitwise AND operation of two numbers.

Input

The first line contains a single integer n(1 ≤ n ≤ 10⁶). The second line contains n integers a₁, a₂, ..., a_n(0 ≤ a_i ≤ 10⁶).

Output

Output a single integer representing the number of required groups modulo 1000000007(10⁹ + 7).

Sample Input

Input

3
2 3 3

Output

0

Input

4
0 1 2 3

Output

10

Input

6
5 2 0 5 2 1

Output

53

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;

 typedef __int64 LL;
 const LL mod=;
 const int maxn=<<;
 int dp[][<<];
 int a[<<];

 LL pow_mod(LL a,LL b)
 {
     LL ret=;a%=mod;
     while(b)
     {
         if(b&) ret=ret*a%mod;
         a=a*a%mod;
         b>>=;
     }
     return ret;
 }

 int main()
 {
     int n,i,j;
     int ans,t,tt;
     scanf("%d",&n);
     for(i=;i<=n;i++) scanf("%d",&a[i]);
     memset(dp,,sizeof(dp));
     for(i=;i<=n;i++)
     {
         if(a[i]&)
         {
             dp[][ a[i] ]++;dp[][ a[i]^ ]++;
         }
         else dp[][ a[i] ]++;
     }
     for(i=;i<;i++)
         for(j=;j<maxn;j++)
         {
             t=<<(i+);
             if(j&t)
             {
                 dp[i+][j]+=dp[i][j];dp[i+][ j-t ]+=dp[i][j];
             }
             else dp[i+][j]+=dp[i][j];
         }
     ans=;
     for(j=;j<maxn;j++)
     {
         t=;
         for(i=;i<;i++)
             if((<<i)&j)
                 t=-t;
         tt=pow_mod(,dp[][j]);
         tt--;
         ans=((ans+t*tt)%mod+mod)%mod;
     }
     printf("%d\n",ans);
     return ;
 }