Problem: You have a cross-section paper with W x H squares, and each of them is painted either in white or black. You want to re-arrange the squares into a neat checkered pattern, in which black and white squares are arranged alternately both in horizontal and vertical directions (the figure shown below is a checkered patter with W = 5 and H = 5). To achieve this goal, you can perform the following two operations as many times you like in an arbitrary sequence: swapping of two arbitrarily chosen columns, and swapping of two arbitrarily chosen rows.

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Create a program to determine, starting from the given cross-section paper, if you can re-arrange them into a checkered pattern.



Input

The input is given in the following format.


W H
c1,1 c1,2 ... c1,W
c2,1 c2,2 ... c2,W
:
cH,1 cH,2 ... cH,W


The first line provides the number of squares in horizontal direction W (2≤W≤1000) and those in vertical direction H(2≤H≤1000). Each of subsequent H lines provides an array of W integers ci,j corresponding to a square of i-th row and j-th column. The color of the square is white if ci,j is 0, and black if it is 1.

Output

Output "yes" if the goal is achievable and "no" otherwise.

Examples

Input

3 2
1 1 0
0 0 1


Output

yes


Input

2 2
0 0
1 1


Output

no

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<set>
#include<queue>
#include<stack>
#include<bitset>
#include<functional>
#include<map>
#include<unordered_set>
using namespace std;
//int p = 998244353;
int p = 1000000007;
#define vel vector<int>
#define vvel vector<vel>
#define rep(i,n) for(int i=0;i<n;i++)
#define sor(v) sort(v.begin(),v.end())
#define mmax(a,b) a=max(a,b)
#define mmin(a,b) a=min(a,b)
#define mkp make_pair
#define pin pair<int,int>
#define qin pair<int,pin>
#define V vector
#define Endl endl
#define veb vector<bool>
#define rev(s) reverse(s.begin(),s.end())
int max_kai = 150000;
vel kai(max_kai, 1);
vel inv_kai;
int rui(int a, int n) {
	if (n == 0) { return 1; }
	int x = rui(a, n / 2);
	x *= x; x %= p;
	if (n % 2 == 1) { x *= a; x %= p; }
	return x;
}
int inv(int a) { return rui(a, p - 2); }
void make_kai() {
	for (int i = 1; i < max_kai; i++) { kai[i] = kai[i - 1] * i; kai[i] %= p; }
	inv_kai = kai;
	rep(i, max_kai) { inv_kai[i] = inv(kai[i]); }
}
int nCr(int n, int r) {
	int a = kai[n] * inv_kai[r]; a %= p;
	a *= inv_kai[n - r]; a %= p;
	return a;
}
vel pa;
int root(int x) {
	if (pa[x] == -1) { return x; }
	int ans = root(pa[x]); pa[x] = ans;
	return ans;
}
void marge(int x, int y) {
	x = root(x);
	y = root(y);
	if (x != y) {pa[x] = y;}
}
#define lower(h,val) lower_bound(h.begin(),h.end(),val)-h.begin()
#define upper(h,val) upper_bound(h.begin(),h.end(),val)-h.begin()
int num_lp(int a, vel &nex) {
	unordered_set<int> se;
	int ans = 0;
	while (true) {
		if (se.find(a) != se.end()) {
			break;
		}
		se.insert(a);
		a = nex[a];
		ans++;
	}
	return ans;
}
int find_cr(int a, vel &nex) {
	unordered_set<int> se;
	while (true) {
		if (se.find(a) != se.end()) {break;}
		se.insert(a);
		a = nex[a];
	}
	return num_lp(a, nex);
}
int inf = 2000000000;
vvel dp;
int solve(vvel &sum, int i, int j, bool is_f) {
	int h = sum.size(); h--;
	int w = sum[0].size();
	if (i == h || j == w) { return 0; }
	if (dp[i][j] == inf) {
		int ans1 = solve(sum, i + 1, j, !is_f);
		int ans2 = solve(sum, i, j + 1, !is_f);
		ans2 += sum[i][j];
		if (is_f) {
			dp[i][j] = min(ans1, ans2);
		}
		else {
			dp[i][j] = max(ans1, ans2);
		}
	}
	return dp[i][j];
}
int rui10(int a) {
	if (a == 0) { return 1; }
	return 10 * rui10(a - 1);
}
int gcd(int a, int b) {
	if (a < b) { swap(a, b); }
	if (b == 0) { return a; }
	return gcd(b, a%b);
}
vvel sol(vvel x,int h,int w) {
	sor(x);
	vvel y(w, vel(h));
	rep(i, w) {
		rep(j, h) {
			y[i][j] = x[j][i];
		}
	}
	sor(y);
	return y;
}
signed main(){
	int w, h; cin >> w >> h;
	vvel x(h, vel(w));
	rep(i, h) {
		rep(j, w) { cin >> x[i][j]; }
	}
	vvel a = sol(x, h, w);
	vvel y(h, vel(w));
	vvel z(h, vel(w));
	rep(i, h) {
		rep(j, w) {
			y[i][j] = (i + j) % 2;
			z[i][j] = 1 - y[i][j];
		}
	}
	y = sol(y, h, w);
	z = sol(z, h, w);
	if (a == y || a == z) {
		cout << "yes" << endl;
	}
	else {
		cout << "no" << endl;
	}
	return 0;
}