Problem:
Score : 100 points

Problem Statement

You are given integers A and B, each between 1 and 3 (inclusive).

Determine if there is an integer C between 1 and 3 (inclusive) such that A \times B \times C is an odd number.

Constraints

All values in input are integers.

1 \leq A, B \leq 3

Input

Input is given from Standard Input in the following format:

A B

Output

If there is an integer C between 1 and 3 that satisfies the condition, print Yes; otherwise, print No.

Sample Input 1

3 1

Sample Output 1

Yes

Let C = 3. Then, A \times B \times C = 3 \times 1 \times 3 = 9, which is an odd number.

Sample Input 2

1 2

Sample Output 2

No

Sample Input 3

2 2

Sample Output 3

No

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

#include <bits/stdc++.h>
using namespace std;
 
#define DEBUG if(1)
#define MAX INT_MAX
#define MAXLL LLONG_MAX
#define MAXU ULLONG_MAX
#define MIN -2000000
#define endl "\n"
#define INF 99999999
#define MOD 1000000007
#define s(n) scanf("%d", &n)
#define ss(a,b) scanf("%d %d",&a,&b)
#define pb push_back
#define mp make_pair
#define M_PI 3.14159265358979323846
#define sz(a) int(a.size())
#define lli long long int
#define rep(i,a,n) for (int i=a;i<n;i++)
#define ler(a,n,vec) for(int i=0;i<n;i++)s(a), vec.pb(a)
typedef vector<lli> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<string, pair<int, char> > ps;
#define DEBUG if(1)
#define F first
#define S second
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int ddx[] = {1, 0};
int ddy[] = {1, 1};


int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  int a, b;
  cin >> a >> b;

  for(int i=a;i<=b;i++){
  	if((a*i*b)%2!=0){
  		cout << "Yes" << endl;
  		return 0;
  	}
  }

cout << "No" << endl;
  return 0;
}