//  zukonit14 - Kunal Raut

#include<bits/stdc++.h>
using namespace std;

// #include "ext/pb_ds/assoc_container.hpp"
// #include "ext/pb_ds/tree_policy.hpp"
// using namespace __gnu_pbds;

// template<class T>
// using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update> ;

// template<class key, class value, class cmp = std::less<key>>
// using omap = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k)  returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;

/*/-----------------------------DEFINES----------------------------------/*/

#define ll                  long long
#define ld                  long double
#define pb                  push_back
#define mp                  make_pair
#define fi                  first
#define se                  second
#define pi                  3.1415926535
#define all(v)              v.begin(), v.end()
#define allr(v)             v.rbegin(), v.rend()
#define ms(s, n)            memset(s, n, sizeof(s))
#define prec(n)             fixed<<setprecision(n)
#define forci(p,n)          for(ll i=p;i<(ll)n;i++)
#define forcj(p,n)          for(ll j=p;j<(ll)n;j++)
#define forc(i,p,n)            for(ll i=p;i<(ll)n;i++)
#define bolt                ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define bits(a)             __builtin_popcountll(a)
#define djokovic            freopen("input00.txt", "r", stdin);freopen("output00.txt", "w", stdout);
#define inrange(i,a,b)      ((i>=min(a,b)) && (i<=max(a,b)))
#define sz(a)               (ll)a.size()
#define sum(a)              ( accumulate ((a).begin(), (a).end(), 0ll))
#define mine(a)             (*min_element((a).begin(), (a).end()))
#define maxe(a)             (*max_element((a).begin(), (a).end()))
#define mini(a)             ( min_element((a).begin(), (a).end()) - (a).begin())
#define maxi(a)             ( max_element((a).begin(), (a).end()) - (a).begin())
#define lowb(a, x)          ( lower_bound((a).begin(), (a).end(), (x)) - (a).begin())
#define uppb(a, x)          ( upper_bound((a).begin(), (a).end(), (x)) - (a).begin())
clock_t time_p              = clock();
void rtime()                {time_p = clock() - time_p; cout << "\nTime Taken : " << fixed << (float)(time_p) / CLOCKS_PER_SEC << "s\n";}


/*/-----------------------------INLINE FUNCTIONS----------------------------------/*/

inline ll gcd(ll a, ll b)          {if (b == 0) return a;    return gcd(b, a % b);}
inline ll lcm(ll a, ll b)          {return (a / gcd(a, b) * b);}
inline bool isprime(ll n)            {ll i; for (i = 2; i <= sqrt(n); i++) {if (n % i == 0)return false;} return true;}
inline ll ceil(ll num, ll den)      {return ((num + den - 1) / den);}

/*/-----------------------------TRACE FUNCTIONS----------------------------------/*/

template<class T> ostream& operator<<(ostream &os, string V) {
    os << "[ ";
    for (auto v : V) os << v << " ";
    return os << "]";
}
template<class T> ostream& operator<<(ostream &os, vector<T> V) {
    os << "[ ";
    for (auto v : V) os << v << " ";
    return os << "]";
}
template<class T> ostream& operator<<(ostream &os, set<T> S) {
    os << "{ ";
    for (auto s : S) os << s << " ";
    return os << "}";
}
template<class T> ostream& operator<<(ostream &os, multiset<T> S) {
    os << "{ ";
    for (auto s : S) os << s << " ";
    return os << "}";
}
template<class T, class T1> ostream& operator<<(ostream &os, map<T, T1> S) {
    os << "{ ";
    for (auto s : S) os << "(" << s.first << "," << s.second << ") ";
    return os << "}";
}
template<class L, class R> ostream& operator<<(ostream &os, pair<L, R> P) {
    return os << "(" << P.first << "," << P.second << ")";
}
#define tracearr(a,n)  {cout << #a<<" : ";cout<<"[ ";for (ll i = 0; i < n; i++) cout << a[i] << " ";cout << "]\n";}
#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;
}
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

/*/-----------------------------MODULAR ARITHMETIC----------------------------------/*/

// const ll modx = 1e9 + 7;
const ll modx = 998244353;
const ll mod = modx;
inline ll add(ll x, ll y, ll mod = modx) {
    x += y;
    if (x >= mod) return x - mod;
    return x % mod;
}
inline ll sub(ll x, ll y, ll mod = modx) {
    x -= y;
    if (x < 0) return x + mod;
    return x;
}
inline ll mul(ll x, ll y, ll mod = modx) {
    return (x * 1ll * y) % mod;
}
inline ll expo(ll x, ll y, ll mod = modx) {
    ll ans = 1;
    while (y) {
        if (y & 1) ans = mul(ans, x, mod);
        x = mul(x, x, mod);
        y >>= 1;
    }
    return ans % mod;
}
inline ll inv(ll x, ll mod = modx) {
    return expo(x, mod - 2, mod);
}
/*/-----------------------------CODE BEGINS----------------------------------/*/
ll rr[] = {0, 1, 1, 1, 0, -1, -1, -1};
ll cc[] = {1, 1, 0, -1, -1, -1, 0, 1};  //x-y -> URDL , row-column -> RDLU; 0-based;

const ll MAXN = 3e5 + 100, MAX2N = 2e3 + 10, LG = 31, INF = 2e18;
const ll N = MAXN;
ll n, m, k, x, q, y, z, a[MAXN];
void solve()
{
    cin >> n >> m >> k;
    if ((n * m - 1) == k)
        cout << "YES\n";
    else
        cout << "NO\n";
}
signed main()
{
    bolt;
#ifndef ONLINE_JUDGE
    djokovic;
#endif
    ll t;
    t = 1;
    cin >> t;
    for (ll i = 1; i <= t; i++)
    {
        // cout << "Case #" << i << ": ";
        solve();
    }
    //rtime();
}

### 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
//  zukonit14 - Kunal Raut

#include<bits/stdc++.h>
using namespace std;

// #include "ext/pb_ds/assoc_container.hpp"
// #include "ext/pb_ds/tree_policy.hpp"
// using namespace __gnu_pbds;

// template<class T>
// using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update> ;

// template<class key, class value, class cmp = std::less<key>>
// using omap = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k)  returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;

/*/-----------------------------DEFINES----------------------------------/*/

#define ll                  long long
#define ld                  long double
#define pb                  push_back
#define mp                  make_pair
#define fi                  first
#define se                  second
#define pi                  3.1415926535
#define all(v)              v.begin(), v.end()
#define allr(v)             v.rbegin(), v.rend()
#define ms(s, n)            memset(s, n, sizeof(s))
#define prec(n)             fixed<<setprecision(n)
#define forci(p,n)          for(ll i=p;i<(ll)n;i++)
#define forcj(p,n)          for(ll j=p;j<(ll)n;j++)
#define forc(i,p,n)            for(ll i=p;i<(ll)n;i++)
#define bolt                ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define bits(a)             __builtin_popcountll(a)
#define djokovic            freopen("input00.txt", "r", stdin);freopen("output00.txt", "w", stdout);
#define inrange(i,a,b)      ((i>=min(a,b)) && (i<=max(a,b)))
#define sz(a)               (ll)a.size()
#define sum(a)              ( accumulate ((a).begin(), (a).end(), 0ll))
#define mine(a)             (*min_element((a).begin(), (a).end()))
#define maxe(a)             (*max_element((a).begin(), (a).end()))
#define mini(a)             ( min_element((a).begin(), (a).end()) - (a).begin())
#define maxi(a)             ( max_element((a).begin(), (a).end()) - (a).begin())
#define lowb(a, x)          ( lower_bound((a).begin(), (a).end(), (x)) - (a).begin())
#define uppb(a, x)          ( upper_bound((a).begin(), (a).end(), (x)) - (a).begin())
clock_t time_p              = clock();
void rtime()                {time_p = clock() - time_p; cout << "\nTime Taken : " << fixed << (float)(time_p) / CLOCKS_PER_SEC << "s\n";}


/*/-----------------------------INLINE FUNCTIONS----------------------------------/*/

inline ll gcd(ll a, ll b)          {if (b == 0) return a;    return gcd(b, a % b);}
inline ll lcm(ll a, ll b)          {return (a / gcd(a, b) * b);}
inline bool isprime(ll n)            {ll i; for (i = 2; i <= sqrt(n); i++) {if (n % i == 0)return false;} return true;}
inline ll ceil(ll num, ll den)      {return ((num + den - 1) / den);}

/*/-----------------------------TRACE FUNCTIONS----------------------------------/*/

template<class T> ostream& operator<<(ostream &os, string V) {
    os << "[ ";
    for (auto v : V) os << v << " ";
    return os << "]";
}
template<class T> ostream& operator<<(ostream &os, vector<T> V) {
    os << "[ ";
    for (auto v : V) os << v << " ";
    return os << "]";
}
template<class T> ostream& operator<<(ostream &os, set<T> S) {
    os << "{ ";
    for (auto s : S) os << s << " ";
    return os << "}";
}
template<class T> ostream& operator<<(ostream &os, multiset<T> S) {
    os << "{ ";
    for (auto s : S) os << s << " ";
    return os << "}";
}
template<class T, class T1> ostream& operator<<(ostream &os, map<T, T1> S) {
    os << "{ ";
    for (auto s : S) os << "(" << s.first << "," << s.second << ") ";
    return os << "}";
}
template<class L, class R> ostream& operator<<(ostream &os, pair<L, R> P) {
    return os << "(" << P.first << "," << P.second << ")";
}
#define tracearr(a,n)  {cout << #a<<" : ";cout<<"[ ";for (ll i = 0; i < n; i++) cout << a[i] << " ";cout << "]\n";}
#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;
}
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

/*/-----------------------------MODULAR ARITHMETIC----------------------------------/*/

// const ll modx = 1e9 + 7;
const ll modx = 998244353;
const ll mod = modx;
inline ll add(ll x, ll y, ll mod = modx) {
    x += y;
    if (x >= mod) return x - mod;
    return x % mod;
}
inline ll sub(ll x, ll y, ll mod = modx) {
    x -= y;
    if (x < 0) return x + mod;
    return x;
}
inline ll mul(ll x, ll y, ll mod = modx) {
    return (x * 1ll * y) % mod;
}
inline ll expo(ll x, ll y, ll mod = modx) {
    ll ans = 1;
    while (y) {
        if (y & 1) ans = mul(ans, x, mod);
        x = mul(x, x, mod);
        y >>= 1;
    }
    return ans % mod;
}
inline ll inv(ll x, ll mod = modx) {
    return expo(x, mod - 2, mod);
}
/*/-----------------------------CODE BEGINS----------------------------------/*/
ll rr[] = {0, 1, 1, 1, 0, -1, -1, -1};
ll cc[] = {1, 1, 0, -1, -1, -1, 0, 1};  //x-y -> URDL , row-column -> RDLU; 0-based;

const ll MAXN = 3e5 + 100, MAX2N = 2e3 + 10, LG = 31, INF = 2e18;
const ll N = MAXN;
ll n, m, k, x, q, y, z, a[MAXN];
void solve()
{
    cin >> n >> m >> k;
    if ((n * m - 1) == k)
        cout << "YES\n";
    else
        cout << "NO\n";
}
signed main()
{
    bolt;
#ifndef ONLINE_JUDGE
    djokovic;
#endif
    ll t;
    t = 1;
    cin >> t;
    for (ll i = 1; i <= t; i++)
    {
        // cout << "Case #" << i << ": ";
        solve();
    }
    //rtime();
}

```