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: Accepted)

    #include <iostream>
    #include <vector>
    #include <set>
    #include <algorithm>
     
    #define rep(i,n) for(int i=0;i< (n);++i)
    #define REP(i,n) for(int i=1;i<=(n);++i)
    #define ALL(a) (a).begin(),(a).end()
    using LLI = long long int;
    LLI GCD(LLI a,LLI b){ if(a<b) return GCD(b,a); else if(a%b==0) return b; else return GCD(b,a%b); }
    LLI LCM(LLI a,LLI b){ return (a/GCD(a,b))*b; }
    using VI  = std::vector<LLI>;
    using PI  = std::pair<LLI,LLI>;
    using VVI = std::vector<VI>;
    using VPI = std::vector<PI>;
     
    int main() {
        std::ios::sync_with_stdio(false);
     
        LLI count[1000] = {};
        LLI H,W;
        std::cin >> H >> W;
        std::string S;
        rep(i,H){
            std::cin >> S;
            rep(j,W)
            ++count[S[j]];
        }
        LLI mod_count[4] = {};
        for(int c='a';c<='z';++c) {
            ++mod_count[count[c]%4];
        }
     
        bool flag = false;
        if(H%2==0&&W%2==0){
            flag = mod_count[1] == 0 && mod_count[2] == 0 && mod_count[3] == 0 ;
        } else if(H%2==1&&W%2==1) {
            flag = mod_count[2]+mod_count[3] <= (H/2+W/2) && mod_count[1]+mod_count[3] == 1;
        } else {
            if(H%2==0) std::swap(H,W);
            // H -> odd, W -> even
            flag = mod_count[1] == 0 && mod_count[2] <= W/2 && mod_count[3] == 0 ;
        }
     
        std::cout << (flag?"Yes":"No") << std::endl;
        return 0;
    }