【动态规划】Gym - 100923A - Por Costel and Azerah

azerah.in / azerah.out

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  (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

给你一个序列，问你有多少个子序列的元素和为偶数。

水dp。分别记f(i)为前i位元素和为偶数的子序列数，g(i)为前i位元素和为奇数的子序列数。瞎几把转移一下就行了。

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
#define MOD 1000000007ll
int T,a[1000010],n;
ll f[1000010],g[1000010];
int main()
{
//	freopen("a.in","r",stdin);
	freopen("azerah.in","r",stdin);
	freopen("azerah.out","w",stdout);
	scanf("%d",&T);
	for(;T;--T)
	  {
	  	memset(f,0,sizeof(f));
	  	memset(g,0,sizeof(g));
	  	scanf("%d",&n);
	  	for(int i=1;i<=n;++i)
	  	  scanf("%d",&a[i]);
	  	if(a[1]%2)
	  	  g[1]=1;
	  	else
	  	  f[1]=1;
	  	for(int i=2;i<=n;++i)
	  	  if(a[i]%2)
	  	    {
	  	      f[i]=(g[i-1]+f[i-1])%MOD;
	  	      g[i]=((f[i-1]+g[i-1])%MOD+1ll)%MOD;
	  	    }
	  	  else
	  	    {
	  	      f[i]=((f[i-1]*2ll)%MOD+1ll)%MOD;
	  	      g[i]=(g[i-1]*2ll)%MOD;
	  	    }
	  	printf("%d\n",f[n]);
	  }
	return 0;
}