You are given an array a consisting of n integer numbers.

You have to color this array in k colors in such a way that: 

  * Each element of the array should be colored in some color; 
  * For each i from 1 to k there should be at least one element colored in the i-th color in the array; 
  * For each i from 1 to k all elements colored in the i-th color should be distinct. 



Obviously, such coloring might be impossible. In this case, print "NO". Otherwise print "YES" and any coloring (i.e. numbers c_1, c_2, ... c_n, where 1 ≤ c_i ≤ k and c_i is the color of the i-th element of the given array) satisfying the conditions above. If there are multiple answers, you can print any.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 5000) — the length of the array a and the number of colors, respectively.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 5000) — elements of the array a.

Output

If there is no answer, print "NO". Otherwise print "YES" and any coloring (i.e. numbers c_1, c_2, ... c_n, where 1 ≤ c_i ≤ k and c_i is the color of the i-th element of the given array) satisfying the conditions described in the problem statement. If there are multiple answers, you can print any.

Examples

Input


4 2
1 2 2 3


Output


YES
1 1 2 2


Input


5 2
3 2 1 2 3


Output


YES
2 1 1 2 1


Input


5 2
2 1 1 2 1


Output


NO

Note

In the first example the answer 2~ 1~ 2~ 1 is also acceptable.

In the second example the answer 1~ 1~ 1~ 2~ 2 is also acceptable.

There exist other acceptable answers for both examples.
var tmp = readline().split(' ');
var n = tmp[0];
var k = tmp[1];

var arr = readline().split(' ');
var arr2 = arr.slice();

arr2.sort( function(x, y ) {
    return x - y; // incresing
});
var mx = 0, cnt = 1;

var mp = [];
var mp2 = [];
var mp3 = [];
var ans = [];

mp[arr2[0]] = 1;
mp2[arr2[0]] = [];

for( var i=1; i<arr.length; i++ ) {
    mp[ arr2[i] ] = 1;
    mp3[ i ] = 0;
    
    if( arr2[i] == arr2[i-1] ) cnt++;
    else {
        mx = Math.max(mx, cnt);
        cnt = 1;
    }
}
mx = Math.max(mx, cnt);

if( mx > k ) {
    print("NO");
} else {
    print("YES");
    
    
    arr.forEach( i => {
        
        ans.push( mp[i] );
        mp3[ mp[i] ]++;
        mp[i]++;
        
    });
    
    var ind = 0;
    
    for( var i=1; i<=k; i++ ) {
        if( !mp3[i] ) {
            //write( i + ' ' + mp3[i] + '\n');
            for( var j=ind; j<n; j++ ) {
                if( mp3[ ans[j] ] > 1 ) {
                    mp3[ ans[j] ]--;
                    ans[j] = i;
                    break;
                }
                
            }
        }
    }
    ans.forEach( i => {
        write( i + ' ');
    });
}


