Problem:
Score : 100 points

Problem Statement

It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board.

The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns.

How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?

Constraints

1 \leq H, W \leq N \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
H
W

Output

Print the answer.

Sample Input 1

3
2
3

Sample Output 1

2

There are two ways to put the notice, as follows:

### ...
### ###
... ###

Here, # represents a square covered by the notice, and . represents a square not covered.

Sample Input 2

100
1
1

Sample Output 2

10000

Sample Input 3

5
4
2

Sample Output 3

8

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <map>
#include <memory>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <random>
#include <tuple>
#include <iomanip>
#include <cstring>

using namespace std;

typedef long long ll;
//typedef pair<int,int> Pint;
//typedef pair<ll, ll> P;
typedef pair<ll, pair<ll, ll>> P;
//typedef tuple<int,int,int> T;
typedef vector<ll> vec;
typedef vector<vec> mat;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define revrep(i, n) for(ll i = (n-1); i >= 0; i--)
#define pb push_back
#define f first
#define s second
const ll INFL = 1LL << 60;//10^18 = 2^60
const int INF = 1 << 30;//10^9
const ll MOD = 1e9 + 7;
//ll MOD  = 998244353;
vector<ll> dy = {0, 0, 1, -1, 1, 1, -1, -1, 0};
vector<ll> dx = {1, -1, 0, 0, 1, -1, 1, -1, 0};

ll max(ll a, ll b){return (a > b) ? a : b;}
ll min(ll a, ll b){return (a < b) ? a : b;}
ll pow_long(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2) res *= x;
    x *= x;
    k /= 2;
  }
  return res;
}
ll pow_mod(ll x, ll k){
  x %= MOD; x += MOD; x %= MOD;
  ll res = 1;
  while(k > 0){
    if(k % 2){
      res *= x; res %= MOD;
    }
    x *= x; x %= MOD;
    k /= 2;
  }
  return res;
}
ll inverse(ll x){return pow_mod(x, MOD - 2);};

//最大公約数
ll gcd(ll a, ll b){
    if(b == 0) return a;
    return gcd(b, a % b);
}

//最小公倍数
ll lcm(ll x, ll y){return x / gcd(x, y) * y;};


ll N, H, W;
int main(void){
   cin >> N >> H >> W;
   cout << (N-H+1) * (N-W+1) << endl;
}