#define _USE_MATH_DEFINES
#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <utility>
#include <complex>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <deque>
#include <tuple>
#include <bitset>
#include <limits>
#include <algorithm>
#include <array>
#include <random>
#include <complex>
#include <regex>
using namespace std;
typedef long double ld;
typedef long long ll;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef complex<ld> compd;
#define quickIO()	{cin.tie(0);	cout.sync_with_stdio(false);}
#define reach(i,a)	for(auto i:a)
#define rep(i,n)	for(int i=0;i<((int)n);i++)
#define REP(i,n)	for(int i=0;i<=((int)n);i++)
#define srep(i,a,n)	for(int i=a;i<((int)n);i++)
#define SREP(i,a,n)	for(int i=a;i<=((int)n);i++)
#define rrep(i,n)	for(int i=n-1;i>=0;i--)
#define RREP(i,n)	for(int i=n;i>=0;i--)
#define all(a)	(a).begin(),(a).end()
#define mp(a,b)	make_pair(a,b)
#define mt	make_tuple
#define pb	push_back
template<typename T> istream & operator >> (istream & is, vector<T> & vec) {
	for (T& x : vec)	is >> x;
	return is;
}
template<typename T> ostream& operator << (ostream & os, vector<T> & vec) {
	os << "[";
	rep(i, vec.size())	os << (i ? ", " : "") << vec[i];
	os << "]";
	return os;
}
template<typename T> istream & operator >> (istream & is, pair<T, T> & p) {
	is >> p.first >> p.second;
	return is;
}
template<typename T> ostream& operator << (ostream & os, pair<T, T> & p) {
	os << p.first << " " << p.second;
	return os;
}
int bitcnt(ll x) {
	x = ((x & 0xAAAAAAAAAAAAAAAA) >> 1) + (x & 0x5555555555555555);
	x = ((x & 0xCCCCCCCCCCCCCCCC) >> 2) + (x & 0x3333333333333333);
	x = ((x & 0xF0F0F0F0F0F0F0F0) >> 4) + (x & 0x0F0F0F0F0F0F0F0F);
	x = ((x & 0xFF00FF00FF00FF00) >> 8) + (x & 0x00FF00FF00FF00FF);
	x = ((x & 0xFFFF0000FFFF0000) >> 16) + (x & 0x0000FFFF0000FFFF);
	x = ((x & 0xFFFFFFFF00000000) >> 32) + (x & 0x00000000FFFFFFFF);
	return x;
}
int bitcnt(int x) {
	x = ((x & 0xAAAAAAAA) >> 1) + (x & 0x55555555);
	x = ((x & 0xCCCCCCCC) >> 2) + (x & 0x33333333);
	x = ((x & 0xF0F0F0F0) >> 4) + (x & 0x0F0F0F0F);
	x = ((x & 0xFF00FF00) >> 8) + (x & 0x00FF00FF);
	x = ((x & 0xFFFF0000) >> 16) + (x & 0x0000FFFF);
	return x;
}
ll sqrtll(ll x) {
	ll left = 0, right = x;
	rep(i, 100) {
		ll mid = (left + right) >> 1;
		if (mid * mid <= x)	left = mid;
		else	right = mid;
	}
	return left;
}
ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a % b);
}
#define debug(x)	printf("Case #%d: ", x)
#define DEBUG 0
const ld infl = 1e100;
const ll mod = 1e9 + 7;
const ld eps = 1e-9;
const ll inf = 1e18;
const int dx[] = { 1,0,-1,0,0 };
const int dy[] = { 0,1,0,-1,0 };

int main() {
	int h, w;	cin >> h >> w;
	vector<string> s(h);	cin >> s;
	int cnt = 0;
	rep(i, h)	rep(j, w) {
		if (s[i][j] == 'o') cnt += ((i + j) & 1 ? 1 : -1);
	}
	cout << (cnt % 3 == 0 ? "YES" : "NO") << endl;
	return 0;
}

/*
白黒は市松模様
1つのみの場合
	除けば勝ちなので必敗
2つの場合
	先に石を除くと負け。ってことでひたすら両者移動を選び続けるが、白黒1つずつなら勝ち
3つの場合
	黒黒黒or白白白で勝ち
4つの場合
	黒黒白白で勝ち
5つの場合
	黒黒黒黒白、黒白白白白で勝ち
6つの場合
	黒黒黒黒黒黒、黒黒黒白白白、白白白白白白で勝ち
ここまでで、黒の個数-白の個数が3の倍数なら勝ちっぽい
*/