Reorder x, y, and z such that x <= y <= z. It is clear that doing so does not affect the answer. Consider the value max(a, b, c). Clearly both y and z has to be equal to this value (because two of the max expressions include it). Therefore, if y != z, the answer is NO. Otherwise (if y = z), output a = x, b = x, and c = z.

Clearly each value from 1 to n occurs exactly twice in the sequence a. Delete the second occurrences of each value from the sequence a to get the answer.

Suppose that in an array, there exists three elements a_i-1, a_i, and a_i+1 such that a_i-1 > a_i < a_i+1 (in other words, there is a local minimum). If such 3 elements exist, the array is not good, because either a_i-1 or a_i+1 must be taken before a_i, violating the requirement that c be non-decreasing. If such 3 elements do not exist, the array is always good (in fact, a good array is always composed of a nondecreasing sequence followed by a nonincreasing sequence; simply take the lowest element at either end each time to obtain a nondecreasing sequence). Insert elements from right to left until the array is no longer good.

Precalculate prefix sums in order to allow counting the number of letters of a certain type in a particular interval in O(1). Do dynamic programming on all intervals that could be involved in the definition of a c-good string (the set of such intervals has the same structure as a segment tree; there are only O(n) of these intervals). dp[i][j] represents the minimum cost to change interval i into a j-good string. To make transitions, decide which half is filled with the same character and which half is a j+1-good string.

If the directed edges already form a cycle, the answer is NO. Otherwise, do a topological sort considering only the directed edges. Then direct the undirected edges from the node that comes before in the topological sort, to the node that comes after.

If the degree of a node is not equal to 1 (mod k), it can never be deleted (because an operation preserves the degree (mod k) of all non-leaf nodes). Call any node which is impossible to delete a blocked node. Any node which lies on a simple path between any two blocked nodes (in other words, any node located in the minimum connected subgraph containing blocked nodes) is also a blocked node, because they can never be turned into leaves. Except in the special cases where there are no blocked nodes at all, or if k = 1 (in these cases, we can always keep deleting until there are no more than k nodes left), it can be shown that all nodes not directly connected to a blocked node can be deleted (by defining a recursive "turn-into-leaf" function); nodes directly connected to a blocked node might also be deletable in groups of k if there are k or more such nodes attached to the same blocked node.

Clearly each value from 1 to n must occur exactly twice in the table in order for the answer to exist. Construct a graph where each node represents a column of the table. Imagine coloring the nodes black and white, where black indicates that the column should be swapped, and white indicates that the column should remain the same. For each value, we want the two occurrences of it to be located in different rows (this condition is both necessary and sufficient). Each value thus imposes a constraint on the two columns where the value can be found; either they must have the same color (black and white, as described above), or they must have different colors. Consider each constraint of this form as an edge. For each "same color" constraint, merge the two nodes at the ends. Then test whether the graph formed by the "different color" constraints is a bipartite graph. If it is not bipartite, the answer is NO; otherwise, for each connected component, choose the smaller part of each bipartite component to be black.