Luntik came out for a morning stroll and found an array a of length n. He calculated the sum s of the elements of the array (s= ∑_{i=1}^{n} a_i). Luntik calls a subsequence of the array a nearly full if the sum of the numbers in that subsequence is equal to s-1.

Luntik really wants to know the number of nearly full subsequences of the array a. But he needs to come home so he asks you to solve that problem!

A sequence x is a subsequence of a sequence y if x can be obtained from y by deletion of several (possibly, zero or all) elements.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The next 2 ⋅ t lines contain descriptions of test cases. The description of each test case consists of two lines.

The first line of each test case contains a single integer n (1 ≤ n ≤ 60) — the length of the array.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9) — the elements of the array a.

Output

For each test case print the number of nearly full subsequences of the array.

Example

Input


5
5
1 2 3 4 5
2
1000 1000
2
1 0
5
3 0 2 1 1
5
2 1 0 3 0


Output


1
0
2
4
4

Note

In the first test case, s=1+2+3+4+5=15, only (2,3,4,5) is a nearly full subsequence among all subsequences, the sum in it is equal to 2+3+4+5=14=15-1.

In the second test case, there are no nearly full subsequences.

In the third test case, s=1+0=1, the nearly full subsequences are (0) and () (the sum of an empty subsequence is 0).
#include <bits/stdc++.h>
using namespace std;

#define ll long long int
#define ld long double
#define vi vector<int>
#define vll vector<long long int>
#define vb vector<bool>
#define vs vector<string>
#define vvi vector<vector<int>>
#define vvll vector<vector<ll>>
#define pll pair<ll, ll>
#define pii pair<int, int>
#define mii map<int, int>
#define mll map<ll, ll>
#define adfor(i, lst) for (auto &i : lst)
#define forEach(i, ar) for (auto i : ar)
#define ifor(i, a, b) for (ll i = a; i < b; i++)
#define ifor1(i, a, b) for (ll i = a; i <= b; i++)
#define dfor(i, a, b) for (ll i = a; i >= b; --i)
#define dfor1(i, a, b) for (ll i = a; i > b; --i)
#define all(v) v.begin(), v.end()
#define maxi INT_MAX
#define mini INT_MIN

const int mod = 1e9 + 7, inf = 998244353, N = 2e5 + 7;

ll gcd(vll &a, ll n)
{
    ll ans = a[0];
    ifor(i, 1, n)
    {
        ans = __gcd(ans, a[i]);
    }
    return ans;
}

int testCases()
{
    int t = 1;
    cin >> t;
    return t;
}

void solve()
{
    ll n;
    cin >> n;
    vll a(n);
    ll sum = 0;
    adfor(i, a)
    {
        cin >> i;
        sum += i;
    }
    if (sum == 1)
    {
        ll ct = count(all(a), 0ll);
        ct++;
        cout << ct << endl;
        return;
    }
    ll ct1 = count(all(a), 1ll);
    ll ct0 = count(all(a), 0ll);
    ll ans = ct1 + ct1 * ct0 + (ct0 > 1);
    cout << ans << endl;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t = testCases();
    while (t--)
    {
        solve();
#ifndef ONLINE_JUDGE
        cout
            << "------------------------- " << endl;
#endif
    }

    return 0;
}
