const int LG = 21;
const int FN = 400005;
const long long MOD = 1e9 + 7;
const long long INF = 1e9;
const long long INFLL = 1e18;

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;

#define forn(i, n) for (int (i) = 0; (i) != (n); (i)++)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define popcount(x) __builtin_popcount(x)
#define popcountll(x) __builtin_popcountll(x)
#define fi first
#define se second
#define re return
#define pb push_back
#define uniq(x) sort(all(x)); (x).resize(unique(all(x)) - (x).begin())

#ifdef LOCAL
#define dbg(x) cerr << __LINE__ << " " << #x << " " << x << endl
#define ln cerr << __LINE__ << endl
#else
#define dbg(x) void(0)
#define ln void(0)
#endif // LOCAL

int cx[4] = {-1, 0, 1, 0};
int cy[4] = {0, -1, 0, 1};

ll inq(ll x, ll y)
{
    if (!y) re 1 % MOD;
    ll l = inq(x, y / 2);
    if (y % 2) re l * l % MOD * x % MOD;
    re l * l % MOD;
}

ll rev(ll x)
{
    return inq(x, MOD - 2);
}

bool __precomputed_combinatorics = 0;
vector<ll> __fact, __ufact, __rev;

void __precompute_combinatorics()
{
    __precomputed_combinatorics = 1;
    __fact.resize(FN);
    __ufact.resize(FN);
    __rev.resize(FN);
    __rev[1] = 1;
    for (int i = 2; i < FN; i++) __rev[i] = MOD - __rev[MOD % i] * (MOD / i) % MOD;
    __fact[0] = 1, __ufact[0] = 1;
    for (int i = 1; i < FN; i++) __fact[i] = __fact[i - 1] * i % MOD, __ufact[i] = __ufact[i - 1] * __rev[i] % MOD;
}

ll fact(int x)
{
    if (!__precomputed_combinatorics) __precompute_combinatorics();
    return __fact[x];
}

ll cnk(int n, int k)
{
    if (k < 0 || k > n) return 0;
    if (!__precomputed_combinatorics) __precompute_combinatorics();
    return __fact[n] * __ufact[n - k] % MOD * __ufact[k] % MOD;
}

int Root(int x, vector<int> &root)
{
    if (x == root[x]) return x;
    return root[x] = Root(root[x], root);
}

void Merge(int v, int u, vector<int> &root, vector<int> &sz)
{
    v = Root(v, root), u = Root(u, root);
    if (v == u) return;
    if (sz[v] < sz[u])
    {
        sz[u] += sz[v];
        root[v] = u;
    }
    else
    {
        sz[v] += sz[u];
        root[u] = v;
    }
}

int ok(int x, int n)
{
    return 0 <= x && x < n;
}

const int N = 400000;

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll n, m;
    cin >> n >> m;
    vector<string> s(n);
    forn(i, n) cin >> s[i];
    vector<int> cnt(26);
    forn(i, n) forn(j, m) cnt[s[i][j] - 'a']++;
    ll r1 = 0, r2 = 0, r4 = 0;
    for (int i = 0; i < 26; i++)
    {
        if (cnt[i] % 4 == 0) r4++;
        else if (cnt[i] % 2 == 0) r2++;
        else r1++;
    }
    if (n % 2 == 1 && m % 2 == 1)
    {
        if (r1 == 1 && r2 <= (n - 1) / 2 + (m - 1) / 2)
        {
            cout << "Yes";
        }
        else
        {
            cout << "No";
        }
        exit(0);
    }
    if (n % 2 == 0 && m % 2 == 0)
    {
        if (r1 == 0 && r2 == 0)
        {
            cout << "Yes";
        }
        else
        {
            cout << "No";
        }
        exit(0);
    }
    if (r1 > 0) cout << "No", exit(0);
    if (n % 2 == 1)
    {
        if (r2 <= m / 2) cout << "Yes";
        else cout << "No";
        return 0;
    }
    else
    {
        if (r2 <= n / 2) cout << "Yes";
        else cout << "No";
        return 0;
    }
}

/* Note:
Check constants at the beginning of the code.
    N is set to 4e5 but be careful in problems with large constant factor.
    Setting N in every problem is more effective.
Check corner cases.
    N = 1
No def int long long for now.
Add something here.
*/


Measured memory_bytes: 256