Problem:
﻿Score: 100 points

Problem Statement

There is a farm whose length and width are A yard and B yard, respectively. A farmer, John, made a vertical road and a horizontal road inside the farm from one border to another, as shown below: (The gray part represents the roads.)

What is the area of this yard excluding the roads? Find it.

Note

It can be proved that the positions of the roads do not affect the area.

Constraints

A is an integer between 2 and 100 (inclusive).

B is an integer between 2 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

A B

Output

Print the area of this yard excluding the roads (in square yards).

Sample Input 1

2 2

Sample Output 1

1

In this case, the area is 1 square yard.

Sample Input 2

5 7

Sample Output 2

24

In this case, the area is 24 square yards.

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

//#include <tourist>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <string>
#include <iomanip>
#include <limits>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <list>
#include <queue>
#include <tuple>
#include <deque>
#include <algorithm>
#include <numeric>
#include <iomanip>
#include <utility>
#include <complex>
#include <functional>
using namespace std;
const int MOD = 1000000007;
typedef long long ll;
typedef pair<ll, ll> p;
const int INF = (1 << 28);
const int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const int Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1}, Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define INF 2e9
#define ALL(v) v.begin(), v.end()
//setprecision(15)有効数字15桁
//-std=c++14
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
struct edge {
    int to, cost;
};
int n, b, c;
vector<int> a;
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> b>>c;
    cout << b*c-b-c+1<< "\n";
}