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

32 3 3

Output

0

Input

40 1 2 3

Output

10

Input

65 2 0 5 2 1

Output

53

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5
 6 typedef __int64 LL;
 7 const LL mod=1000000007;
 8 const int maxn=1<<20;
 9 int dp[20][1<<20];
10 int a[1<<20];
11
12 LL pow_mod(LL a,LL b)
13 {
14     LL ret=1;a%=mod;
15     while(b)
16     {
17         if(b&1) ret=ret*a%mod;
18         a=a*a%mod;
19         b>>=1;
20     }
21     return ret;
22 }
23
24 int main()
25 {
26     int n,i,j;
27     int ans,t,tt;
28     scanf("%d",&n);
29     for(i=1;i<=n;i++) scanf("%d",&a[i]);
30     memset(dp,0,sizeof(dp));
31     for(i=1;i<=n;i++)
32     {
33         if(a[i]&1)
34         {
35             dp[0][ a[i] ]++;dp[0][ a[i]^1 ]++;
36         }
37         else dp[0][ a[i] ]++;
38     }
39     for(i=0;i<19;i++)
40         for(j=0;j<maxn;j++)
41         {
42             t=1<<(i+1);
43             if(j&t)
44             {
45                 dp[i+1][j]+=dp[i][j];dp[i+1][ j-t ]+=dp[i][j];
46             }
47             else dp[i+1][j]+=dp[i][j];
48         }
49     ans=0;
50     for(j=0;j<maxn;j++)
51     {
52         t=1;
53         for(i=0;i<20;i++)
54             if((1<<i)&j)
55                 t=-t;
56         tt=pow_mod(2,dp[19][j]);
57         tt--;
58         ans=((ans+t*tt)%mod+mod)%mod;
59     }
60     printf("%d\n",ans);
61     return 0;
62 }

codeforces 449D DP+容斥