You have a fair six-sided die with positive integer face values $a_1,a_2,\ldots,a_6$. Construct another fair six-sided die with positive integer face values $b_1,b_2,\ldots,b_6$ whose sum is exactly $k$. Face values may repeat on either die.

Roll each die once, independently. The new die wins if and only if its value is strictly greater than the original die's value; a tie is not a win. Determine whether you can make it win at least $19$ of the $36$ equally likely pairs of faces (that is, with probability strictly greater than $50\%$). If so, construct such a die.

Input

The first line contains six integers $a_1,a_2,\ldots,a_6$ ($1\le a_i\le 10^9$), the labels of the original die.

The second line contains an integer $k$ ($1\le k\le 10^{18}$), the required sum of the new die's labels.

Output

If no valid new die exists, print NO.

Otherwise, print YES on the first line and six positive integers $b_1,b_2,\ldots,b_6$ on the second line. Their sum must be exactly $k$, and the new die must win at least $19$ of the $36$ pairs of faces.

The labels may appear in any order. If multiple answers exist, print any one. The strings YES and NO are case-sensitive.

Examples

Input 1

1 2 3 4 5 6
25

Output 1

YES
1 1 2 7 7 7

Input 2

1 1 1 1 1 1
9

Output 2

NO

Note

For the first example, another valid choice of labels is $5,4,4,4,4,4$. These labels sum to $5+4+4+4+4+4=25$. Its face labeled $5$ beats four faces of the original die, and each face labeled $4$ beats three. Thus, it wins $4+5\cdot3=19$ pairs.

In the second example, a face must be labeled at least $2$ to win. Each such face wins six pairs, so at least four are needed to win $19$ pairs. Even if the other two faces are labeled $1$, the sum is at least $2+2+2+2+1+1=10>9$.