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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var i,j:longint;
    n,m,num2,num4:longint;
    ch:char;
    num:array['a'..'z']of longint;
    have:longint;
begin
  readln(n,m);
  for i:=1 to n do
  begin
    for j:=1 to m do
    begin
      read(ch);
      inc(num[ch]);
    end;
    readln;
  end;
  if (n mod 2=1)and(m mod 2=1) then num4:=((n-1)div 2) * ((m-1) div 2) else
  if (n mod 2=0)and(m mod 2=1) then num4:=(  n  div 2) * ((m-1) div 2) else
  if (n mod 2=1)and(m mod 2=0) then num4:=((n-1)div 2) * (  m   div 2) else
  if (n mod 2=0)and(m mod 2=0) then num4:=(  n  div 2) * (  m   div 2);
  for i:=97 to 122 do
  if (num[char(i)]>=4)and(num4>0) then
  begin
    have:=num[char(i)]div 4;
    if have>num4 then
    begin
      num4:=0;
      dec(num[char(i)],num4*4);
    end else
    begin
      dec(num4,have);
      dec(num[char(i)],have*4);
    end;
  end;
  if num4>0 then
  begin
    writeln('No');
    exit;
  end;
  if (n mod 2=1)and(m mod 2=1) then num2:=(n-1) div 2+(m-1) div 2 else
  if (n mod 2=0)and(m mod 2=1) then num2:=m div 2 else
  if (n mod 2=1)and(m mod 2=0) then num2:=n div 2 else
  if (n mod 2=0)and(m mod 2=0) then num2:=0; 
  for i:=97 to 122 do
  if (num[char(i)]>=2)and(num2>0) then
  begin
    have:=num[char(i)] div 2;
    if have>num2 then num2:=0 else dec(num2,have);
  end;
  if num2>0 then
  begin
    writeln('No');
    exit;
  end;
  writeln('Yes');
end.