The Eight Puzzle, among other sliding-tile puzzles, is one of the famous problems in artificial intelligence. Along with chess, tic-tac-toe and backgammon, it has been used to study search algorithms.

The Eight Puzzle can be generalized into an M × N Puzzle where at least one of M and N is odd. The puzzle is constructed with MN ? 1 sliding tiles with each a number from 1 to MN ? 1 on it packed into a M by N frame with one tile missing. For example, with M = 4 and N = 3, a puzzle may look like:

Let‘s call missing tile 0. The only legal operation is to exchange 0 and the tile with which it shares an edge. The goal of the puzzle is to find a sequence of legal operations that makes it look like:

The following steps solve the puzzle given above.

Given an M × N puzzle, you are to determine whether it can be solved.

The input consists of multiple test cases. Each test case starts with a line containing M and N (2 ≤ M, N ≤ 999). This line is followed by M lines containing N numbers each describing an M × N puzzle.

The input ends with a pair of zeroes which should not be processed.

Output one line for each test case containing a single word YES if the puzzle can be solved and NO otherwise.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 2e9
#define met(a,b) memset(a,b,sizeof a)
typedef long long ll;
using namespace std;
const int N = 999*999+5;
const int M = 4e5+5;
int n,m,tot=0,cnt=0;
int head[N],ans[N];
int tree[N];
int a[N];
void add(int k,int num) {
    while(k<=999*999-1) {
        tree[k]+=num;
        k+=k&(-k);
    }
}
int Sum(int k) {
    int sum=0;
    while(k>0) {
        sum+=tree[k];
        k-=k&(-k);
    }
    return sum;
}
int main() {

    while(scanf("%d%d",&n,&m),n||m) {
        int x,y,t,s=0,nu=0;
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++) {
                scanf("%d",&t);
                if(t==0)
                    x=i,y=j;
                else
                    a[nu++]=t;
            }
       met(tree,0);
        for(int i=nu-1; i>=0; i--) {
            s+=Sum(a[i]-1);
            add(a[i],1);
        }
        if(m&1)
            if(s&1)puts("NO");
            else   puts("YES");
        else if(((n-x)^s)&1) puts("NO");
        else            puts("YES");
    }
    return 0;
}