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 (5.3.2) (judge verdict: Accepted)

local h, w = io.read("*n", "*n", "*l")
local words = {}
for i = 1, h do
  local str = io.read()
  for j = 1, w do
    local sstr = str:sub(j, j)
    if not words[sstr] then words[sstr] = 1 else words[sstr] = words[sstr] + 1 end
  end
end
local a, b = 0, 0
for k, v in pairs(words) do
  if v % 2 == 1 then
    if b == 0 then
      b = v
    else
      b = -1
    end
  elseif v % 4 == 2 then
    a = a + 1
  end
end

if h % 2 == 0 and w % 2 == 0 then
  if a == 0 and b == 0 then
    print("Yes")
  else
    print("No")
  end
elseif h % 2 == 1 and w % 2 == 1 then
  if 0 < b then
    if (b - 1) % 4 == 2 then
      a = a + 1
    end
    local hh, hw = (h - 1) // 2, (w - 1) // 2
    if a <= hh + hw then
      print("Yes")
    else
      print("No")
    end
  else
    print("No")
  end
else
  if b ~= 0 then
    print("No")
  else
    if h % 2 == 1 then h, w = w, h end
    local hh = h // 2
    if a <= hh then
      print("Yes")
    else
      print("No")
    end
  end
end