#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define FORR2(x,y,arr) for(auto& [x,y]:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
template<class T> bool chmax(T &a, const T &b) { if(a<b){a=b;return 1;}return 0;}
template<class T> bool chmin(T &a, const T &b) { if(a>b){a=b;return 1;}return 0;}
//-------------------------------------------------------

int A[8][8];
bitset<100000200> dp;
int p10[9];
int B[10];
ll mask[10],NG;
void solve() {
	int i,j,k,l,r,x,y; string s;
	
	p10[0]=1;
	FOR(i,8) p10[i+1]=p10[i]*10;
	
	FOR(y,8) {
		cin>>s;
		FOR(x,8) {
			A[y][x]=s[x]-'0';
			B[y]+=A[y][x]*p10[x];
			mask[y]|=(6LL+A[y][x])<<(5*x);
		}
		NG|=1LL<<(5*y+4);
	}
	dp[0]=1;
	queue<int> Q;
	Q.push(0);
	while(Q.size()) {
		int cur=Q.front();
		Q.pop();
		int nex=cur/10+(cur%10)*10000000;
		if(dp[nex]==0) {
			dp[nex]=1;
			Q.push(nex);
		}
		ll V=0;
		FOR(i,8) V+=(1LL*cur/p10[i]%10)<<(5*i);;
		for(x=1;x<8;x++) {
			nex=cur+B[x];
			ll V2=V+mask[x];
			
			if((V2&NG)==0&&dp[nex]==0) {
				dp[nex]=1;
				Q.push(nex);
			}
		}
	}
	
	if(dp[B[0]]) {
		cout<<"Yes"<<endl;
	}
	else {
		cout<<"No"<<endl;
	}
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}