A - Por Costel and Azerah

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-emptysubsequences 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  (the size of the array) and the second line will contain  integers  (), separated by single spaces.

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

题意：给你一个序列，问你有多少个子序列的元素和为偶数。输入第一行为测试样例数，每行样例两行，第一行为序列的长度，第二行为序列。

思路：对于每一位可以选择取或不取，对于奇数,dp结果为上一位的奇数情况和偶数情况之和，对于偶数，则是偶数情况是偶数情况*2,并且奇数情况是上一位奇数情况*2，

#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <iostream>
#define mod 1000000007
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

int t,n;
ll x,dp[1000010][2];
int main()
{
    freopen("azerah.in","r",stdin);
    freopen("azerah.out","w",stdout);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        dp[0][1]=1;//1代表位偶数时
        dp[0][0]=0;//0代表位奇数时
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&x);
            if(x%2==0)
            {
                dp[i][1]=dp[i-1][1]*2%mod;
                dp[i][0]=dp[i-1][0]*2%mod;
            }
            else
            {
                dp[i][1]=dp[i][0]=(dp[i-1][1]+dp[i-1][0])%mod;
            }
        }
        printf("%lld\n",dp[n][1]-1);
    }
}