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>
#include <ext/pb_ds/assoc_container.hpp>
#include <bitset>
using namespace __gnu_pbds;
using namespace std;

#define ff first
#define ss second
#define int long long
#define float long double
#define endl '\n'
#define pb push_back
#define pii pair<int, int>
#define vi vector<int>
#define mii map<int, int>
#define umii unordered_map<int, int>
#define pqb priority_queue<int>
#define pqs priority_queue<int, vi, greater<int>>
#define mod 1000000007
#define inf INT_MAX
#define ninf INT_MIN
#define sz(x) ((int)(x).size())
#define rep(i, n) for (int i = 0; i < (n); i++)
#define vrep(x, v) for (auto x : v)
#define all(n) n.begin(), n.end()
#define prec(n, i) cout << fixed << setprecision(n) << i << '\n';
#define print(array) vrep(i, array) { cout << i << ' '; }cout << endl;
#define printMat(memo) vrep(i, memo) {vrep(j, i) cout << j << ' ';cout << endl;}

typedef tree<pii, null_type, less<pii>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

struct pair_hash
{
    inline std::size_t operator()(const std::pair<int, int> &v) const
    {
        return v.first * 31 + v.second;
    }
};

int pow(int a, int p, int m = mod)
{
    int res = 1;
    while (p > 0)
    {
        if (p & 1)
            res = (res * a) % m;
        a = (a * a) % m;
        p >>= 1;
    }
    return res;
}

void c_p_c()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
}

int32_t main()
{
    c_p_c();
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vi v(n);
        rep(i, n)cin >> v[i];
        int sum = 0, oneCnt = 0, zeroCnt = 0;
        vrep(i, v) {
            sum += i;
            if (i == 1)oneCnt++;
            if (i == 0)zeroCnt++;
        }
        // cout << oneCnt << ' ' << (sum == 0) << ' ' << zeroCnt << endl;
        int res = (1 << zeroCnt) * oneCnt;
        cout << res << endl;
    }
    return 0;
}