#include<bits/stdc++.h>
#define scn(n) scanf("%d",&n)

#define lscn(n) scanf("%lld ",&n)

#define scn2(n1,n2) scanf("%d%d",&n1,&n2)

#define lscn2(n1,n2) scanf("%lld%lld",&n1,&n2)

#define lpri(n) printf("%lld ",n)

#define lpri2(n1,n2) printf("%lld %lld\n",n1,n2)

#define pri(n) printf("%d ",n)

#define pri2(n1,n2) printf("%d %d\n",n1,n2)

#define pln() printf("\n")

#define priln(n) printf("%d\n",n)

#define lpriln(n) printf("%lld\n",n)

#define rep(i,init,n) for(int i=init;i<n;i++)

#define repv(i,init,n) for(int i=init;i>=n;i--)


#define mp make_pair

#define code lli n,k;cin>>n>>k;lli a[n];loop0(i,n) cin>>a[i]
#define F first

#define S second
#define lli long long int
#define all(x) x.begin(), x.end()
#define toh_shuru_karte_hai_bina_kise_bakchodi_ke ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
#define test lli t;cin>>t;while(t--)
#define arrin lli n;cin>>n;lli a[n];for(lli i=0;i<n;i++)cin>>a[i]       // a%b=a-b*floor(a/b);
#define floop for(lli i=0;i<n;i++)
#define mod 1000000007
#define defvec vector<lli>v
#define ub upper_bound
#define pb push_back
#define bs binary_search
#define INF ll(8230219982008199610)
#define MINF ll(-8230219982008199610)
#define INT_BITS 16
#define ones(n) __builtin_popcountll(n)
#define focc(n) __builtin_ffsll(n) //to count index of first 1 from right in binary representation of a number starting from index 1
#define leadingzero(n) __builtin_clzll(n)
#define trailingzeroes(n) __builtin_ctzll(n)
//string s=R"(HIMANS&\nSHAN\t)"; we can print a whole string same as given by doing this
//vector<int>v={1,2,3,4};
//vector<int>w=move(v);
//after this v={};
//w={1,2,3,4};
// to reverse a string reverse(str.begin(), str.end());
//  OR
// string rev = string(str.rbegin(),str.rend());
#define loop0(i,n) for(int i=0; i<n; i++)
#define loop1(i,n) for(int i=1; i<=n; i++)
#define rev0(i,n)  for(int i=n-1; i>=0; i--)
#define rev1(i,n)  for(int i=n; i>0; i--)
using ll=long long;
//using namespace boost::multiprecision ;
using namespace std;
//All 4 digit palindromic numbers are divisible by 11.
//A number of form pow(2,n) has exactly N+1 divisors
//If we repeat a three-digit number twice, to form a six-digit number. The result will will be divisible by 7, 11 and 13, and dividing by all three will give your original three-digit number.
bool isvalid(lli x,lli y,lli n,lli m)
{
    if(x<0||x>(n-1)||y<0||y>(m-1))
    return false;
    //if(visited[x][y]==true)
    //return false;
    return true;
}
lli bs(lli a[],lli start,lli end,lli x,lli n)
{
    lli mid=start+(end-start)/2;
    if(start>end)
    return -1;
    if(a[mid]==x)
    return mid;
    else if(a[mid]>x)
    return bs(a,start,mid-1,x,n);
    else 
    return bs(a,mid+1,end,x,n);
}      
int main()
{
 toh_shuru_karte_hai_bina_kise_bakchodi_ke;
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
test{
   lli n,m,k;
   cin>>n>>m>>k;
   if(n%2!=0&&m%2!=0&&k>1)
   cout<<"NO\n";
   else
   {
       lli c=1;
       while(n%2==0)
       {
          c=c*2;
          n=n/2;
       }
       while(m%2==0)
       {
           c=c*2;
           m=m/2;
       }
       if(k>c)
       cout<<"NO\n";
       else
       cout<<"YES\n";
   }
}
}