//SAGAR SINGH
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
// #pragma GCC optimization("unroll-loops")
#include<bits/stdc++.h>
using namespace std;
#define fastio ios_base::sync_with_stdio;cin.tie(NULL);cout.tie(0);
#define ll long long 
#define ld long double 
#define rep(i,a,b) for(ll i=a;i<b;i++) 
#define repr(i,a,b) for(ll i=a;i>=b;i--)
#define vll vector<long long>
#define vpl vector<pair<ll,ll> >
#define sll set<long long>
#define msll multiset<long long>
#define mll map<ll,ll>
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define in(v) for(auto &x:v) cin>>x;
#define out(v) for(auto &x:v) {cout<<x<<" ";} cout<<'\n';
#define str string
#define pb  push_back
#define ins insert
#define lb lower_bound
#define ub upper_bound
#define ff first
#define ss second
/*--- bit-stuff ----*/
#define setbits(a) ( __builtin_popcountll(a))
#define right(a) (__builtin_ctzll(a))
#define left(a) (__builtin_clzll(a))
#define parity(a) (__builtin_parityll(a))

#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <typename Arg1>
    void __f(const char* name, Arg1&& arg1){
        cout << name << " : " << arg1 << endl;
        //use cerr if u want to display at the bottom
    }
    template <typename Arg1, typename... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args){
        const char* comma = strchr(names + 1, ','); cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
    }
#else
#define trace(...)
#endif


const ll mod = 1e9+7;
const ll N = 1e5+1;
const ll INF = 1e18;


void solve(){

ll n,m,k;cin>>n>>m>>k;
cout<<(k==((n-1)+(m-1)*n)? "YES\n":"NO\n");


}

int main(){
  ios_base::sync_with_stdio;
  cin.tie(NULL);cout.tie(0);
  ll t=1;
  cin>>t;
  while(t--) 
    solve();

  return 0;
}

### Prompt

In CPP, your task is to solve 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
//SAGAR SINGH
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
// #pragma GCC optimization("unroll-loops")
#include<bits/stdc++.h>
using namespace std;
#define fastio ios_base::sync_with_stdio;cin.tie(NULL);cout.tie(0);
#define ll long long 
#define ld long double 
#define rep(i,a,b) for(ll i=a;i<b;i++) 
#define repr(i,a,b) for(ll i=a;i>=b;i--)
#define vll vector<long long>
#define vpl vector<pair<ll,ll> >
#define sll set<long long>
#define msll multiset<long long>
#define mll map<ll,ll>
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define in(v) for(auto &x:v) cin>>x;
#define out(v) for(auto &x:v) {cout<<x<<" ";} cout<<'\n';
#define str string
#define pb  push_back
#define ins insert
#define lb lower_bound
#define ub upper_bound
#define ff first
#define ss second
/*--- bit-stuff ----*/
#define setbits(a) ( __builtin_popcountll(a))
#define right(a) (__builtin_ctzll(a))
#define left(a) (__builtin_clzll(a))
#define parity(a) (__builtin_parityll(a))

#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <typename Arg1>
    void __f(const char* name, Arg1&& arg1){
        cout << name << " : " << arg1 << endl;
        //use cerr if u want to display at the bottom
    }
    template <typename Arg1, typename... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args){
        const char* comma = strchr(names + 1, ','); cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
    }
#else
#define trace(...)
#endif


const ll mod = 1e9+7;
const ll N = 1e5+1;
const ll INF = 1e18;


void solve(){

ll n,m,k;cin>>n>>m>>k;
cout<<(k==((n-1)+(m-1)*n)? "YES\n":"NO\n");


}

int main(){
  ios_base::sync_with_stdio;
  cin.tie(NULL);cout.tie(0);
  ll t=1;
  cin>>t;
  while(t--) 
    solve();

  return 0;
}

```