or the New Year, Polycarp decided to send postcards to all his n friends. He wants to make postcards with his own hands. For this purpose, he has a sheet of paper of size w×h, which can be cut into pieces.
Polycarp can cut any sheet of paper w×h that he has in only two cases:
If w is even, then he can cut the sheet in half and get two sheets of size w2×h;
If h is even, then he can cut the sheet in half and get two sheets of size w×h2;
If w and h are even at the same time, then Polycarp can cut the sheet according to any of the rules above.
After cutting a sheet of paper, the total number of sheets of paper is increased by 1.
Help Polycarp to find out if he can cut his sheet of size w×h at into n or more pieces, using only the rules described above.
Input
The first line contains one integer t (1≤t≤104) — the number of test cases. Then t test cases follow.
Each test case consists of one line containing three integers w, h, n (1≤w,h≤104,1≤n≤109) — the width and height of the sheet Polycarp has and the number of friends he needs to send a postcard to.
Output
For each test case, output on a separate line:
“YES”, if it is possible to cut a sheet of size w×h into at least n pieces;
“NO” otherwise.
You can output “YES” and “NO” in any case (for example, the strings yEs, yes, Yes and YES will be recognized as positive).
Example
inputCopy
5
2 2 3
3 3 2
5 10 2
11 13 1
1 4 4
outputCopy
YES
NO
YES
YES
YES
Note
In the first test case, you can first cut the 2×2 sheet into two 2×1 sheets, and then cut each of them into two more sheets. As a result, we get four sheets 1×1. We can choose any three of them and send them to our friends.
In the second test case, a 3×3 sheet cannot be cut, so it is impossible to get two sheets.
In the third test case, you can cut a 5×10 sheet into two 5×5 sheets.
In the fourth test case, there is no need to cut the sheet, since we only need one sheet.
In the fifth test case, you can first cut the 1×4 sheet into two 1×2 sheets, and then cut each of them into two more sheets. As a result, we get four sheets 1×1.
长和宽是偶数就可以切一刀多两倍个数出来，将这些切法乘起来比较n的个数大小
#include<map>
#include<stack>
#include<queue>
#include<string>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define ls (k<<1)
#define rs (k<<1|1)
#define pb push_back
#define mid ((l+r)>>1)
#include<bits/stdc++.h>
using namespace std;
const int p=1e4+7;
const int mod=1e9+7;
const int maxn=1e6+1;
typedef long long ll;
const int inf=0x3f3f3f3f;
void solve(){
int t;
cin>>t;
while(t--){
int w,h,n;
cin>>w>>h>>n;
int ans=1;
int tmp=1;
while(w){
if(w%2==0){
w/=2;
tmp*=2;
}
else
break;
}
ans*=tmp;
tmp=1;
while(h){
if(h%2==0)
{
h/=2;
tmp*=2;
}
else
break;
}
ans*=tmp;
if(ans>=n){
cout<<"YES"<<endl;
}
else{
cout<<"NO"<<endl;
}
}
}
signed main()
{
ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
solve();
return 0;
}