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 Kotlin (1.3.71) (judge verdict: Accepted)

fun main(arr: Array<String>) {
    println(if(getAns()) "Yes" else "No")
}

fun getAns():Boolean {
    val (h,w) = readLine()!!.split(" ").map { it.toInt() }
    val map = (1..h).map { readLine()!! }
    val cnt = mutableMapOf<Char, Int>()
    map.forEach { line -> line.forEach { cnt[it] = (cnt[it]?:0)+1 } }
    if(h%2==0) {
        if(w%2==0) {
            return cnt.values.all { it%4==0 }
        } else {
            if(cnt.values.any{it%2!=0}) {
                return false
            }
            return cnt.values.filter { it%4!=0 && it%2==0 }.count()*2 <= h
        }
    } else {
        if(w%2==0) {
            if(cnt.values.any{it%2!=0}) {
                return false
            }
            return cnt.values.filter { it%4!=0 && it%2==0 }.count()*2 <= w
        } else {
            if(cnt.values.count { it%2!=0 } != 1) {
                return false
            }
            return cnt.values.filter { it%4!=0&&it%2==0 }.count()*2 <= w+h-2
        }
    }
}