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 Lua (Lua 5.3.5) (judge verdict: Wrong Answer)

local h,w=io.read("n","n","l")
local map={}
for i=1,h do
    local s=io.read()
    for j=1,w do
        local a=s:sub(j,j)
        map[a]=(map[a] or 0)+1
    end
end

local count={}
for k,v in pairs(map) do
    count[v]=(count[v] or 0)+1
end

local g1=(h%2)*(w%2)
local g2=(h%2)*(w//2)+(w%2)*(h//2)
local g4=(h//2)*(w//2)

while g1>0 do
    local checker=false
    for i=1,10000 do
        if i%4==1 or i%4==3 then
            if count[i] and count[i]>0 then
                count[i]=count[i]-1
                count[i-1]=(count[i-1] or 0)+1
                checker=true
                g1=g1-1
            end
        end
    end
    if not checker then
        print("No")
        return
    end
end

while g4>0 do
    local checker=false
    for i=1,10000 do
        if i%2==0 then
            if count[i] and count[i]>0 then
                count[i]=count[i]-1
                count[i-4]=(count[i-4] or 0)+1
                checker=true
                g4=g4-1
            end
        end
    end
    if not checker then
        print("No")
        return
    end
end

while g2>0 do
    local checker=false
    for i=1,10000 do
        if i%2==0 then
            if count[i] and count[i]>0 then
                count[i]=count[i]-1
                count[i-2]=(count[i-2] or 0)+1
                checker=true
                g2=g2-1
            end
        end
    end
    if not checker then
        print("No")
        return
    end
end

print("Yes")