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 (GCC 5.4.1) (judge verdict: Output Limit Exceeded)

/*

結果:
*/
#define _USE_MATH_DEFINES
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <limits.h>  
#include <stdbool.h>   
#define inf (INT_MAX-1)
#define INF 9223372036854775807
#define PI 3.14159265358979323846;
#define EPS 1e-10
#define sq(n) ((n)*(n))
#define rep(i,n) for(i=0;i<n;i++)
#define rev(i,n) for(i=n-1;i>=0;i--)
#define sort(a,n) qsort(a,n,sizeof(TYPE),cmp)
#define sort_r(a,n) qsort(a,n,sizeof(TYPE),cmp_r)
#define chsort(s,n) qsort(s,n,sizeof(char),cmp)
#define chsort_r(s,n) qsort(s,n,sizeof(char),char_cmp_r)
#define TYPE long long
#define ll long long
#define MEMSET(a) memset(a,0,sizeof(a))
#define MEMSET_U(a) memset(a,-1,sizeof(a))
const int mod = (int)1e09 + 7;

//#define DEBUG1
//#define DEBUG2
//#define DEBUGF
#define DUMMY

int in(void) { int i; scanf("%d", &i); return i; }
long long llin(void) { long long i; scanf("%lld", &i); return i; }
double din(void) { double i; scanf("%lf", &i); return i; }
void chin(char s[]) { scanf("%s", s); }

void print(int a) { printf("%d\n", a); }
void llprint(long long a) { printf("%lld\n", a); }
void dprint(double a) { printf("%.10f\n", a); }
void print2(int a, int b) { printf("%d %d\n", a, b); }
/*
int max(int a, int b) { if (a>b) { return a; }return b; }
int min(int a, int b) { if (a<b) { return a; }return b; }
*/
long long llmax(long long a, long long b) { return a > b ? a : b; }
long long llmin(long long a, long long b) { return a < b ? a : b; }
double dmax(double a, double b) { return a > b ? a : b; }
int cmp(const void *a, const void *b) { return *(TYPE *)a - *(TYPE *)b; }
int cmp_r(const void *a, const void *b) { return *(TYPE *)b - *(TYPE *)a; }
int char_cmp(const void *a, const void *b) { return strcmp((char *)a, (char *)b); }
int char_cmp_r(const void *a, const void *b) { return strcmp((char *)b, (char *)a); }
void swap(int *a, int *b) { int t = *a; *a = *b; *b = t; }




int main() {
	int A = in();
	int B = in();
	int c;

	for (int i = 1; i = 3; i++) {
		if ((A*B*c) % 2) {
			printf("%s\n", "Yes");
		}
		else {
			printf("%s\n", "No");
		}
	}




#ifdef DEBUGF
	getch();
#endif
	return 0;
	}