Por Costel the Pig has received a royal invitation to the palace of the Egg-Emperor of Programming, Azerah. Azerah had heard of the renowned pig and wanted to see him with his own eyes. Por Costel, having arrived at the palace, tells the Egg-Emperor that he looks "tasty". Azerah feels insulted (even though Por Costel meant it as a compliment) and, infuratied all the way to his yolk, threatens to kill our round friend if he doesn't get the answer to a counting problem that he's been struggling with for a while

Given an array of numbers, how many non-empty subsequences of this array have the sum of their numbers even ? Calculate this value mod  (Azerah won't tell the difference anyway)

Help Por Costel get out of this mischief!

Input

The file azerah.in will contain on its first line an integer , the number of test cases. Each test has the following format: the first line contains an integer 

It is guaranteed that the sum of  over all test cases is at most 

Output

The file azerah.out should contain  lines. The -th line should contain a single integer, the answer to the -th test case.

Example

Input

2
3
3 10 1
2
4 2

Output

3
3

題意：給你一個序列，問你有多少個子序列的元素和為偶數。

直接看程式碼就懂，用到了全排的問題

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
int mod=1e9+7;
int main()
{
     freopen("azerah.in","r",stdin);
    freopen("azerah.out","w",stdout);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        long long n;
        cin>>n;
        long long l=0,r=0,x;
        for(int i=1; i<=n; i++)
        {
            cin>>x;
            if(x%2==0)
                l++;
            else r++;
        }
        long long ans=1,sum=1;
        for(int i=1; i<=l; i++)
        {
            ans*=2;
            ans%=mod;
        }
        for(int i=1; i<r; i++)
        {
            sum*=2;
            sum%=mod;
        }
        ans--;
        sum--;
        cout<<(sum+ans+sum*ans)%mod<<endl;
    }
    return 0;
}