#include <bits/stdc++.h>
using namespace std;
#define FAST ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define int               long long
#define pb                push_back
#define ppb               pop_back
#define pf                push_front
#define ppf               pop_front
#define all(x)            (x).begin(),(x).end()
#define rall(x)           x.rbegin(), x.rend()
#define uniq(v)           (v).erase(unique(all(v)),(v).end())
#define sz(x)             (int)((x).size())
#define fr                first
#define sc                second
#define pii               pair<int,int>
#define rep(i,a,b)        for(int i=a;i<b;i++)
#define repi(i,a,b)       for(ll i=b-1;i>=a;i--)
#define mem1(a)           memset(a,-1,sizeof(a))
#define mem0(a)           memset(a,0,sizeof(a))
#define ppc               builtin_popcount
#define ppcll             builtin_popcountll
#define sz5               100005
#define sz6               1000005
#define V(a)              vector<a>
#define MP               make_pair
#define MT               make_tuple
#define G(a,b)           get<a>(b)
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const int N = 5e5 + 5;
template<typename T1, typename T2>istream& operator>>(istream& in, pair<T1, T2> &a) {in >> a.fr >> a.sc; return in;}
template<typename T1, typename T2>ostream& operator<<(ostream& out, pair<T1, T2> a) {out << a.fr << " " << a.sc; return out;}
template<typename T, typename T1>T amax(T &a, T1 b) {if (b > a)a = b; return a;}
template<typename T, typename T1>T amin(T &a, T1 b) {if (b < a)a = b; return a;}

void solve(void)
{
    
}
signed main()
{
#ifndef ONLINE_JUDGE
    //FOR GETTING INPUT FROM INPUT.TEXT
    freopen("input", "r", stdin);
    //for getting output to output.txt
    freopen("output", "w", stdout);
#endif
    FAST
    int t;
    cin >> t;
    while (t--)
    {
        int n,m,k;
        cin>> n>>m>>k;
        int cost= (n-1)+n*(m-1);
        if(cost==k)
        cout<<"Yes"<<"\n";
        else
        cout<<"No"<<"\n";
    }

    return 0;
}
### Prompt

Please create a solution in Cpp 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 FAST ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define int               long long
#define pb                push_back
#define ppb               pop_back
#define pf                push_front
#define ppf               pop_front
#define all(x)            (x).begin(),(x).end()
#define rall(x)           x.rbegin(), x.rend()
#define uniq(v)           (v).erase(unique(all(v)),(v).end())
#define sz(x)             (int)((x).size())
#define fr                first
#define sc                second
#define pii               pair<int,int>
#define rep(i,a,b)        for(int i=a;i<b;i++)
#define repi(i,a,b)       for(ll i=b-1;i>=a;i--)
#define mem1(a)           memset(a,-1,sizeof(a))
#define mem0(a)           memset(a,0,sizeof(a))
#define ppc               builtin_popcount
#define ppcll             builtin_popcountll
#define sz5               100005
#define sz6               1000005
#define V(a)              vector<a>
#define MP               make_pair
#define MT               make_tuple
#define G(a,b)           get<a>(b)
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const int N = 5e5 + 5;
template<typename T1, typename T2>istream& operator>>(istream& in, pair<T1, T2> &a) {in >> a.fr >> a.sc; return in;}
template<typename T1, typename T2>ostream& operator<<(ostream& out, pair<T1, T2> a) {out << a.fr << " " << a.sc; return out;}
template<typename T, typename T1>T amax(T &a, T1 b) {if (b > a)a = b; return a;}
template<typename T, typename T1>T amin(T &a, T1 b) {if (b < a)a = b; return a;}

void solve(void)
{
    
}
signed main()
{
#ifndef ONLINE_JUDGE
    //FOR GETTING INPUT FROM INPUT.TEXT
    freopen("input", "r", stdin);
    //for getting output to output.txt
    freopen("output", "w", stdout);
#endif
    FAST
    int t;
    cin >> t;
    while (t--)
    {
        int n,m,k;
        cin>> n>>m>>k;
        int cost= (n-1)+n*(m-1);
        if(cost==k)
        cout<<"Yes"<<"\n";
        else
        cout<<"No"<<"\n";
    }

    return 0;
}
```