You are storing an integer array of length m in a database. To maintain internal integrity and protect data, the database stores n copies of this array.

Unfortunately, the recent incident may have altered the stored information in every copy in the database.

It's believed, that the incident altered at most two elements in every copy. You need to recover the original array based on the current state of the database.

In case there are multiple ways to restore the array, report any. If there is no array that differs from every copy in no more than two positions, report that as well.

Input

The first line contains integers n and m (2 ≤ n; 1 ≤ m; n ⋅ m ≤ 250 000) — the number of copies and the size of the array.

Each of the following n lines describes one of the currently stored copies in the database, it consists of m integers s_{i, 1}, s_{i, 2}, ..., s_{i, m} (1 ≤ s_{i, j} ≤ 10^9).

Output

If there is an array consistent with all given copies, print "Yes" and then the array itself. The array must have length m and contain integers between 1 and 10^9 only.

Otherwise, print "No".

If there are multiple possible arrays, print any of them.

Examples

Input


3 4
1 10 10 100
1 1 1 100
10 100 1 100


Output


Yes
1 10 1 100


Input


10 7
1 1 1 1 1 1 1
1 1 1 1 1 1 2
1 1 1 1 1 2 2
1 1 1 1 2 2 1
1 1 1 2 2 1 1
1 1 2 2 1 1 1
1 2 2 1 1 1 1
2 2 1 1 1 1 1
2 1 1 1 1 1 1
1 1 1 1 1 1 1


Output


Yes
1 1 1 1 1 1 1


Input


2 5
2 2 1 1 1
1 1 2 2 2


Output


No

Note

In the first example, the array [1, 10, 1, 100] differs from first and second copies in just one position, and from the third copy in two positions.

In the second example, array [1, 1, 1, 1, 1, 1, 1] is the same as the first copy and differs from all other copies in at most two positions.

In the third example, there is no array differing in at most two positions from every database's copy.
//@formatter:off
#include<bits/stdc++.h>
#define overload4(_1,_2,_3,_4,name,...) name
#define rep1(i,n) for (ll i = 0; i < ll(n); ++i)
#define rep2(i,s,n) for (ll i = ll(s); i < ll(n); ++i)
#define rep3(i,s,n,d) for(ll i = ll(s); i < ll(n); i+=d)
#define rep(...) overload4(__VA_ARGS__,rep3,rep2,rep1)(__VA_ARGS__)
#define rrep1(i,n) for (ll i = ll(n)-1; i >= 0; i--)
#define rrep2(i,n,t) for (ll i = ll(n)-1; i >= (ll)t; i--)
#define rrep3(i,n,t,d) for (ll i = ll(n)-1; i >= (ll)t; i-=d)
#define rrep(...) overload4(__VA_ARGS__,rrep3,rrep2,rrep1)(__VA_ARGS__)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define popcount(x) __builtin_popcount(x)
#define pb push_back
#define eb emplace_back
#ifdef __LOCAL
#define debug(...) { cout << #__VA_ARGS__; cout << ": "; print(__VA_ARGS__); cout << flush; }
#else
#define debug(...) void(0)
#endif
#define INT(...) int __VA_ARGS__;scan(__VA_ARGS__)
#define LL(...) ll __VA_ARGS__;scan(__VA_ARGS__)
#define STR(...) string __VA_ARGS__;scan(__VA_ARGS__)
#define CHR(...) char __VA_ARGS__;scan(__VA_ARGS__)
#define DBL(...) double __VA_ARGS__;scan(__VA_ARGS__)
#define LD(...) ld __VA_ARGS__;scan(__VA_ARGS__)
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<int,int>;
using LP = pair<ll,ll>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using vd = vector<double>;
using vvd = vector<vector<double>>;
using vs = vector<string>;
using vc = vector<char>;
using vvc = vector<vector<char>>;
using vb = vector<bool>;
using vvb = vector<vector<bool>>;
using vp = vector<P>;
using vvp = vector<vector<P>>;
template<class S,class T> istream& operator>>(istream &is,pair<S,T> &p) { return is >> p.first >> p.second; }
template<class S,class T> ostream& operator<<(ostream &os,const pair<S,T> &p) { return os<<'{'<<p.first<<","<<p.second<<'}'; }
template<class T> istream& operator>>(istream &is,vector<T> &v) { for(T &t:v){is>>t;} return is; }
template<class T> ostream& operator<<(ostream &os,const vector<T> &v) { os<<'[';rep(i,v.size())os<<v[i]<<(i==int(v.size()-1)?"":","); return os<<']'; }
template<class T> void vecout(const vector<T> &v,char div='\n') { rep(i,v.size()) cout<<v[i]<<(i==int(v.size()-1)?'\n':div);}
template<class T> bool chmin(T& a,T b) {if(a > b){a = b; return true;} return false;}
template<class T> bool chmax(T& a,T b) {if(a < b){a = b; return true;} return false;}
void scan(){}
template <class Head, class... Tail> void scan(Head& head, Tail&... tail){ cin >> head; scan(tail...); }
template<class T> void print(const T& t){ cout << t << '\n'; }
template <class Head, class... Tail> void print(const Head& head, const Tail&... tail){ cout<<head<<' '; print(tail...); }
template<class... T> void fin(const T&... a) { print(a...); exit(0); }
const string yes[] = {"no","yes"};
const string Yes[] = {"No","Yes"};
const string YES[] = {"NO","YES"};
const int inf = 1001001001;
const ll linf = 1001001001001001001;
//@formatter:on

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    INT(n, m);
    vvi s(n, vi(m));
    cin >> s;
    auto imp = [] {
        fin(Yes[0]);
    };
    auto check = [&](vi &t, int fr = -1) {
        assert(t.size() == m);
        int dt = -1;
        rep(i, n) {
            int cnt = 0;
            rep(j, m) cnt += s[i][j] != t[j];
            if (cnt > 3) return;
            if (cnt == 3) {
                if (fr < 0) return;
                if (dt > 0 and dt != s[i][fr]) return;
                dt = s[i][fr];
            }
        }
        if(dt < 0) dt = 1;
        if(fr >= 0) t[fr] = dt;
        print(Yes[1]);
        vecout(t, ' ');
        exit(0);
    };
    check(s[0]);
    rep(k, 1, n) {
        vi ls;
        rep(j, m) {
            if (s[0][j] != s[k][j]) ls.pb(j);
        }
        if (ls.size() > 4) imp();
        if (ls.size() <= 2) continue;
        if (ls.size() == 3) {
            rep(bit, 1 << 3) {
                if (popcount(bit) == 0 or popcount(bit) == 3) continue;
                vi now = s[0];
                rep(i, 3) if (bit >> i & 1) now[ls[i]] = s[k][ls[i]];
                check(now);
            }
            rep(i, 3) rep(j, 3) {
                    if (i == j) continue;
                    vi now = s[0];
                    now[ls[i]] = s[k][ls[i]];
                    now[ls[j]] = 0;
                    check(now, ls[j]);
                }
        }
        if (ls.size() == 4) {
            rep(bit, 1 << 4) {
                if (popcount(bit) != 2) continue;
                vi now = s[0];
                rep(i, 4) if (bit >> i & 1) now[ls[i]] = s[k][ls[i]];
                check(now);
            }
        }
        imp();
    }
}
