Problem:
Score : 400 points

Problem Statement

We have an H-by-W matrix.
Let a_{ij} be the element at the i-th row from the top and j-th column from the left.
In this matrix, each a_{ij} is a lowercase English letter.

Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A.
Here, he wants to satisfy the following condition:

Every row and column in A' can be read as a palindrome.

Determine whether he can create a matrix satisfying the condition.

Note

A palindrome is a string that reads the same forward and backward.
For example, a, aa, abba and abcba are all palindromes, while ab, abab and abcda are not.

Constraints

1 ≤ H, W ≤ 100

a_{ij} is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}

Output

If Snuke can create a matrix satisfying the condition, print Yes; otherwise, print No.

Sample Input 1

3 4
aabb
aabb
aacc

Sample Output 1

Yes

For example, the following matrix satisfies the condition.

abba
acca
abba

Sample Input 2

2 2
aa
bb

Sample Output 2

No

It is not possible to create a matrix satisfying the condition, no matter how we rearrange the elements in A.

Sample Input 3

5 1
t
w
e
e
t

Sample Output 3

Yes

For example, the following matrix satisfies the condition.

t
e
w
e
t

Sample Input 4

2 5
abxba
abyba

Sample Output 4

No

Sample Input 5

1 1
z

Sample Output 5

Yes

Solution in Scala (2.13.1) (judge verdict: Wrong Answer)

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)
    val H, W = sc.nextInt()
    sc.nextLine()
    val A: Array[Array[Char]] = Array.fill(H)(sc.nextLine().toCharArray)
    // Aの要素を自由に並べ替えて、どの行およびどの列もそれぞれ回文になっているようにすることができるかどうか
    // 実際に行列を作る必要性はなさそう
    // 同じ文字の数が多い順に、回文を作れるかどうかを再帰的に調べていく
    // 現在の行列の縦,横のサイズ*2の数だけ同じ文字が存在すれば,埋める
    // 埋めると,縦を埋めた場合,横のサイズが2縮まる.横を埋めた場合,縦のサイズが2縮まる
    // 両方とも埋めれるということは,縦と横のサイズは同じなので,どちらを選んでも結果は同じになる
    // 同じ数の文字を持つ種類のどちらを先に選んでも,結果は同じになる

    def solve(H: Int, W: Int, A: Array[Array[Char]]): Boolean = {
      val charSizes = A.flatten.groupBy(identity).values.map(_.length).toList.sorted(Ordering.Int.reverse)
      @scala.annotation.tailrec
      def recursive(h: Int, w: Int, charSizes: List[Int]): Boolean =
        charSizes match {
          case Nil => true
          case size :: tail =>
            if (h == 1 && w == 1 && size == 1) true
            else if (h * 2 == size) recursive(h, w - 2, tail)
            else if (w * 2 == size) recursive(h - 2, w, tail)
            else false
        }
      recursive(H, W, charSizes)
    }

    val res = if (solve(H, W, A)) "Yes" else "No"
    println(res)
  }

}