#include <cstdio>
#include <iostream>
#include <algorithm>
#include <complex>
#include <queue>
#include <map>
#include <string>
using namespace std;

#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define FORR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
#define pb push_back
#define ALL(a) (a).begin(),(a).end()

#define PI 3.1415926535

typedef long long ll;
typedef pair<int, int> P;
//typedef complex<double> C;

const int INF = 99999999;
const int CENTER = 10;
const int H = 2 * CENTER + 1;
const int W = 2 * CENTER + 1;

int gx, gy;
int d[H][W];

int dx[8] = {2, 2, 1, 1, -1, -1, -2, -2};
int dy[8] = {1, -1, 2, -2, 2, -2, 1, -1};

void input() {
	cin >> gx >> gy;
}

bool inside(int x, int y) {
    return x >= 0 && x < H && y >= 0 && y < W;
}

int bfs() {
    queue<P> que;
    REP(i, H) REP(j, W) d[i][j] = INF;
    que.push(P(CENTER, CENTER));
    d[CENTER][CENTER] = 0;

    while (!que.empty()) {
        P p = que.front(); que.pop();
        if (p.first == gx && p.second == gy)
             break;

        REP(k, 8) {
            int nx = p.first + dx[k], ny = p.second + dy[k];

            if (inside(nx, ny) && d[nx][ny] == INF) {
                que.push(P(nx, ny));
                d[nx][ny] = d[p.first][p.second] + 1;
            }
        }
    }
    return d[gx][gy];
}

void solve() {
	if (abs(gx) > 6 || abs(gy) > 6) {
		cout << "NO" << endl;
		return;
	}

	gx += CENTER; gy += CENTER;
	int ans = bfs();
	if (ans <= 3) {
		cout << "YES" << endl;
	} else {
		cout << "NO" << endl;
	}

}


int main() {
	input();
	solve();
}