#include "bits/stdc++.h"
using namespace std;
#define en      '\n'
#define endl    '\n'
#define ll      long long signed
#define pb      push_back
#define append  push_back
#define all(v)  (v).begin(),(v).end()
#define FOR(i,a,b,k) for(int i=a; k>0?i<b:i>b; i+=k)
#define fori(a) for(int i=0; i<a; i++)
#define forj(a) for(int j=0; j<a; j++)
#define fork(a) for(int k=0; k<a; k++)
#define MOD     1000000007
#define MAX     1000010
#define mset(a,s,k) memset(a,k,s*sizeof(a[0]))
#define ff      first
#define ss      second
#define __      <<" "<<
#define _       <<en

ll pow_(ll a,ll e,ll m=MOD){ll r=1;while(e){if(e&1)r=r*a%m;a=a*a%m;e>>=1;}return r;} 
int inv(ll a) {return int(pow_(a,MOD-2,MOD));}int add(ll a, ll b){return (a+b+MOD)%MOD;} int mul(ll a, ll b)
{return (MOD+a*b%MOD)%MOD;}int di(ll a, ll b){return mul(a,inv(b));}ll  ceil(ll a,ll b){return a/b+(a%b!=0);}
template<typename T1,typename T2>ostream&operator<<(ostream&O,pair<T1,T2>a){O<<"( ";O<<a.ff<<", "<<a.ss;O<<" )";return O;}
template<typename T>ostream&operator<<(ostream&O,vector<T>V){O<<"[ ";fori(V.size())O<<V[i]<<(i==V.size()-1?"":", ");O<<" ]";return O;}
template<typename T>ostream&operator<<(ostream&O,set<T>S){O<<"{ ";for(T i:S)O<<i<<", ";O<<" }";return O;}
template<typename T1,typename T2>ostream&operator<<(ostream&O,map<T1,T2>S){O<<"{ ";for(auto i:S)O<<i<<", ";O<<" }";return O;}
template<typename T=int>class print{public:vector<T>V;print(vector<T>v){V=v;cout<<" <"<<V<<"> ";}
print(T*x,int n){fori(n)V.push_back(x[i]);cout<<" <"<<V<<"> ";}friend ostream&operator<<(ostream&O,print p){return O;}};
#define int ll
#define vi vector<int>
#define tp pair<int,int>

int arr[MAX]; int N,M,K;
void reset();
map<array<int,3>, int> MEM;

int rec(int x, int y, int s){
    if(s==0 && x==N && y==M) return 1;
    if(x>N || y>M || s<=0) return 0;
    if(MEM.count({x,y,s})) return MEM[{x,y,s}];
    return (MEM[{x,y,s}]=(rec(x,y+1,s-x) || rec(x+1,y,s-y)));
}

int solve(){
    cin>>N>>M>>K;
    MEM.clear();
    return rec(1,1,K);
    // int a=min(N,M);
    // int b=max(N,M);
    // int mx=min(a-1+(b-1)*b, (b-1)+(a-1)*a);
    // int mi=
    // int c=abs(N-M);
    // int a=min(N,M);
    // int b=max(N,M);
    // int mx=max(a-1+(b-1)*b, (b-1)+(a-1)*a);
    // int mi=
    // return 0;
}

void reset(){

    
    // Clear Vectors, maps, sets
    // Set default values of arr
}
 
signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t=1;
    cin>>t;
    while(t--){
        // solve();
        // cout<<solve()<<en;
        cout<<(solve()?"Yes":"No")<<endl;
    }
    return 0;
}


### Prompt

Generate a Cpp solution to the following problem:

There is a n × m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).

You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can: 

  * move right to the cell (x, y + 1) — it costs x burles; 
  * move down to the cell (x + 1, y) — it costs y burles. 



Can you reach cell (n, m) spending exactly k burles?

Input

The first line contains the single integer t (1 ≤ t ≤ 100) — the number of test cases.

The first and only line of each test case contains three integers n, m, and k (1 ≤ n, m ≤ 100; 0 ≤ k ≤ 10^4) — the sizes of grid and the exact amount of money you need to spend.

Output

For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).

Example

Input


6
1 1 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000


Output


YES
NO
YES
NO
YES
NO

Note

In the first test case, you are already in the final cell, so you spend 0 burles.

In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) → (1, 2) → (2, 2) or (1, 1) → (2, 1) → (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.

In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.

### Response

```cpp
#include "bits/stdc++.h"
using namespace std;
#define en      '\n'
#define endl    '\n'
#define ll      long long signed
#define pb      push_back
#define append  push_back
#define all(v)  (v).begin(),(v).end()
#define FOR(i,a,b,k) for(int i=a; k>0?i<b:i>b; i+=k)
#define fori(a) for(int i=0; i<a; i++)
#define forj(a) for(int j=0; j<a; j++)
#define fork(a) for(int k=0; k<a; k++)
#define MOD     1000000007
#define MAX     1000010
#define mset(a,s,k) memset(a,k,s*sizeof(a[0]))
#define ff      first
#define ss      second
#define __      <<" "<<
#define _       <<en

ll pow_(ll a,ll e,ll m=MOD){ll r=1;while(e){if(e&1)r=r*a%m;a=a*a%m;e>>=1;}return r;} 
int inv(ll a) {return int(pow_(a,MOD-2,MOD));}int add(ll a, ll b){return (a+b+MOD)%MOD;} int mul(ll a, ll b)
{return (MOD+a*b%MOD)%MOD;}int di(ll a, ll b){return mul(a,inv(b));}ll  ceil(ll a,ll b){return a/b+(a%b!=0);}
template<typename T1,typename T2>ostream&operator<<(ostream&O,pair<T1,T2>a){O<<"( ";O<<a.ff<<", "<<a.ss;O<<" )";return O;}
template<typename T>ostream&operator<<(ostream&O,vector<T>V){O<<"[ ";fori(V.size())O<<V[i]<<(i==V.size()-1?"":", ");O<<" ]";return O;}
template<typename T>ostream&operator<<(ostream&O,set<T>S){O<<"{ ";for(T i:S)O<<i<<", ";O<<" }";return O;}
template<typename T1,typename T2>ostream&operator<<(ostream&O,map<T1,T2>S){O<<"{ ";for(auto i:S)O<<i<<", ";O<<" }";return O;}
template<typename T=int>class print{public:vector<T>V;print(vector<T>v){V=v;cout<<" <"<<V<<"> ";}
print(T*x,int n){fori(n)V.push_back(x[i]);cout<<" <"<<V<<"> ";}friend ostream&operator<<(ostream&O,print p){return O;}};
#define int ll
#define vi vector<int>
#define tp pair<int,int>

int arr[MAX]; int N,M,K;
void reset();
map<array<int,3>, int> MEM;

int rec(int x, int y, int s){
    if(s==0 && x==N && y==M) return 1;
    if(x>N || y>M || s<=0) return 0;
    if(MEM.count({x,y,s})) return MEM[{x,y,s}];
    return (MEM[{x,y,s}]=(rec(x,y+1,s-x) || rec(x+1,y,s-y)));
}

int solve(){
    cin>>N>>M>>K;
    MEM.clear();
    return rec(1,1,K);
    // int a=min(N,M);
    // int b=max(N,M);
    // int mx=min(a-1+(b-1)*b, (b-1)+(a-1)*a);
    // int mi=
    // int c=abs(N-M);
    // int a=min(N,M);
    // int b=max(N,M);
    // int mx=max(a-1+(b-1)*b, (b-1)+(a-1)*a);
    // int mi=
    // return 0;
}

void reset(){

    
    // Clear Vectors, maps, sets
    // Set default values of arr
}
 
signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t=1;
    cin>>t;
    while(t--){
        // solve();
        // cout<<solve()<<en;
        cout<<(solve()?"Yes":"No")<<endl;
    }
    return 0;
}


```