#include <bits/stdc++.h>
#define F first
#define S second
#define pb push_back

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

#define ordered_set tree<pair<ll,int>, null_type,less<pair<ll,int>>, rb_tree_tag,tree_order_statistics_node_update>

using namespace std;
typedef long long ll;

const int MOD = 1e9 + 7;
const int mxK = 1e5 + 9;
const int mxN = 1e2 + 9;
const int INF = 1e7 + 7;
const double eps = 1e-7;

ll dp1[mxN][mxK], dp2[mxN][mxK], pre1[mxN][mxK], suff2[mxN][mxK], cnt[mxK];

void solve () {
    int N, K;
    cin >> N >> K;

    int n = sqrt (N);
    for (int j = 1; j <= n; j++) {
        int r = N / j;

        int l = ceil (N / (j + 1.0));
        if (l * (j + 1) == N)
            l++;

        cnt[j] = r - l + 1;
        dp1[1][j] = 1;
        dp2[1][j] = cnt[j];

        pre1[1][j] = pre1[1][j - 1] + dp1[1][j];
    }

    for (int j = n; j >= 1; j--)
        suff2[1][j] = suff2[1][j + 1] + dp2[1][j];

    for (int i = 2; i <= K; i++) {
        for (int j = 1; j <= n; j++) {
            dp1[i][j] = suff2[i - 1][j] + pre1[i - 1][n];
            dp2[i][j] = pre1[i - 1][j] * cnt[j];

            pre1[i][j] = pre1[i][j - 1] + dp1[i][j];
        }

        for (int j = n; j >= 1; j--)
            suff2[i][j] = suff2[i][j + 1] + dp2[i][j];
    }

    ll res = 0;
    for (int j = 1; j <= n; j++) {
        //cout << dp1[K][j] << ' ' << dp2[K][j] << "\n";

        res += dp1[K][j] + dp2[K][j];
    }

    cout << res << "\n";
}

int main () {
    int t = 1;
    //cin >> t;

    //preCalc ();
    while (t--)
        solve ();

    return 0;
}

/*
 ((2?3)?8)
 1 1

 *
 *
 * */