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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
using namespace std;

int main(void){

	int n, m, i, j, count[26]={0},nolma[2] ={0},kazu=0;
	char c[100][100];
	
	cin >> n  >> m;
	
	for(i = 0; i < n; i++){
            cin >> c[i];
	}
	for(i = 0; i < n; i++){
	    for(j = 0; j < m;j++){
	        if(c[i][j] == 'a'){
	            count[0]++;
	        }
            else if(c[i][j] == 'b'){
	            count[1]++;
	        }
            else if(c[i][j] == 'c'){
	            count[2]++;
	        }
	        else if(c[i][j] == 'd'){
	            count[3]++;
	        }
            else if(c[i][j] == 'e'){
	            count[4]++;
	        }
	        else if(c[i][j] == 'f'){
	            count[5]++;
	        }
            else if(c[i][j] == 'g'){
	            count[6]++;
	        }
	        else if(c[i][j] == 'h'){
	            count[7]++;
	        }
            else if(c[i][j] == 'i'){
	            count[8]++;
	        }
	        else if(c[i][j] == 'j'){
	            count[9]++;
	        }
            else if(c[i][j] == 'k'){
	            count[10]++;
	        }
	        else if(c[i][j] == 'l'){
	            count[11]++;
	        }
            else if(c[i][j] == 'm'){
	            count[12]++;
	        }
	        else if(c[i][j] == 'n'){
	            count[13]++;
	        }
            else if(c[i][j] == 'o'){
	            count[14]++;
	        }
	        else if(c[i][j] == 'p'){
	            count[15]++;
	        }
            else if(c[i][j] == 'q'){
	            count[16]++;
	        }
	        else if(c[i][j] == 'r'){
	            count[17]++;
	        }
            else if(c[i][j] == 's'){
	            count[18]++;
	        }
	        else if(c[i][j] == 't'){
	            count[19]++;
	        }
            else if(c[i][j] == 'u'){
	            count[20]++;
	        }
	        else if(c[i][j] == 'v'){
	            count[21]++;
	        }
            else if(c[i][j] == 'w'){
	            count[22]++;
	        }
	        else if(c[i][j] == 'x'){
	            count[23]++;
	        }
            else if(c[i][j] == 'y'){
	            count[24]++;
	        }
	        else if(c[i][j] == 'z'){
	            count[25]++;
	        }
	    }
	}

    for(i = 0; i < 26; i++){
        if(count[i] == 1){
            nolma[0]++;
        }
        else if(count[i] == 2){
            nolma[1]++;
        }
        else{
            kazu =  kazu + count[i] / 4;
        }
    }
    if(n % 2 == 1 && m % 2 ==1){
        if(nolma[0] == 1 && nolma[1] <= (n - 1)/2 + (m - 1)/2){
            cout << "Yes" << endl;
        }
        else if(nolma[0] == 1 && kazu >= m * n / 4 - nolma[1]/2){
            cout << "Yes" << endl;
        }else{
            cout << "No" << endl;
        }
    }else if(n % 2 == 0 && m % 2 == 0){
        if(nolma[0] == 0 && nolma[1] == 0)
            cout << "Yes" << endl;
        else 
            cout << "No" << endl;
    }else if(n % 2 == 0 && m % 2 == 1){
        if(nolma[0] == 0 && nolma[1] <= n/2){
            cout << "Yes" << endl;
        }
        else if(nolma[0] == 0 && kazu >= m * n / 4 - nolma[1]/2){
            cout << "Yes" << endl;
        }else{
            cout << "No" << endl;
        }
    }else{
        if(nolma[0] == 0 && nolma[1] <= m/2){
            cout << "Yes" << endl;
        }
        else if(nolma[0] == 0 && kazu >= m * n / 4 - nolma[1]/2){
            cout << "Yes" << endl;
        }
        else 
            cout << "No" << endl;
    }
	return 0;
}