#include<bits/stdc++.h>

using namespace std;

#define ll long long

const int mod=1e9+7;

ll dp[150005];

int a[305];

int main()

{

    int i,j,n,t;

    scanf("%d",&t);

    while(t--){

        scanf("%d",&n);

        ll sum=0;

        for(i=1;i<=n;i++){

            scanf("%d",&a[i]);

            sum+=a[i];

        }

        sort(a+1,a+1+n,greater<int>());

        memset(dp,0,sizeof dp);

        dp[0]=1;

        ll ans=0;

        for(i=1;i<=n;i++){

            for(j=sum;j>=a[i];j--){

                dp[j]=(dp[j]+dp[j-a[i]])%mod;

                if(j>=sum-j&&j-a[i]<=sum-j)ans=(ans+dp[j-a[i]])%mod;

                //满足条件就把选了a【i】的方案数dp[j-a[i]]加上 而dp【j】则包含了没选a【i】的方案数，没选的前面已经加过了

            }

        }

        printf("%lld\n",ans);

    }

    return 0;

}