// Original C++ solution (slower):
#include<iostream>

#include<algorithm>

#include<cmath>

#include<map>

#include<stdio.h>

#include<vector>

using namespace std;

#define int long long

#define rep(s,i,n) for(int i=s;i<n;i++)

#define c(n) cout<<n<<endl;

#define ic(n) int n;cin>>n;

#define sc(s) string s;cin>>s;

#define mod 1000000007

#define inf 1000000000000000007

#define f first

#define s second

#define mini(c,a,b) *min_element(c+a,c+b)

#define maxi(c,a,b) *max_element(c+a,c+b)

int keta(int x){

	rep(0,i,30){

		if(x<10){

			return i+1;

		}

		x=x/10;

	}

}

int gcd(int x,int y){

	int aa=x,bb=y;

	rep(0,i,1000){

		aa=aa%bb;

		if(aa==0){

			return bb;

		}	

		bb=bb%aa;

		if(bb==0){

			return aa;

		}

	}

}	

int lcm(int x,int y){

	int aa=x,bb=y;

	rep(0,i,1000){

		aa=aa%bb;

		if(aa==0){

			return x/bb*y;

		}	

		bb=bb%aa;

		if(bb==0){

			return x/aa*y;

		}

	}

}

char mem[114514];

char t[114514];

char c[114514];

string s[114514];

int a[114514],b[114514];

signed main(){

	ic(n) ic(a)

	c((n-1)/a+1)

}

// Optimized solution to the same problem (faster):
#include<iostream>

#include<algorithm>

#include<cmath>

#include<map>

#include<stdio.h>

#include<vector>

#include<queue>

#include<math.h>

#include<deque>

using namespace std;

#define int long long

#define rep(s,i,n) for(int i=s;i<n;i++)

#define c(n) cout<<n<<endl;

#define ic(n) int n;cin>>n;

#define sc(s) string s;cin>>s;

#define mod 1000000007

#define inf 1000000000000000007

#define f first

#define s second

#define mini(c,a,b) *min_element(c+a,c+b)

#define maxi(c,a,b) *max_element(c+a,c+b)

#define pi 3.141592653589793238462643383279

#define e_ 2.718281828459045235360287471352

#define P pair<int,int>

#define upp(a,n,x) upper_bound(a,a+n,x)-a;

#define low(a,n,x) lower_bound(a,a+n,x)-a;

#define UF UnionFind 

//printf("%.12Lf\n",);

int keta(int x) {

	rep(0, i, 30) {

		if (x < 10) {

			return i + 1;

		}

		x = x / 10;

	}

}

int gcd(int x, int y) {

	if (x == 0 || y == 0)return x + y;

	int aa = x, bb = y;

	rep(0, i, 1000) {

		aa = aa % bb;

		if (aa == 0) {

			return bb;

		}

		bb = bb % aa;

		if (bb == 0) {

			return aa;

		}

	}

}

int lcm(int x, int y) {

	int aa = x, bb = y;

	rep(0, i, 1000) {

		aa = aa % bb;

		if (aa == 0) {

			return x / bb * y;

		}

		bb = bb % aa;

		if (bb == 0) {

			return x / aa * y;

		}

	}

}

bool p(int x) {

	if (x == 1)return false;

	rep(2, i, sqrt(x) + 1) {

		if (x % i == 0 && x != i) {

			return false;

		}

	}

	return true;

}

int max(int a, int b) {

	if (a >= b)return a;

	else return b;

}

string maxst(string s, string t) {

	int n = s.size();

	int m = t.size();

	if (n > m)return s;

	else if (n < m)return t;

	else {

		rep(0, i, n) {

			if (s[i] > t[i])return s;

			if (s[i] < t[i])return t;

		}

		return s;

	}

}

int min(int a, int b) {

	if (a >= b)return b;

	else return a;

}

int n2[41];

int nis[41];

int nia[41];

int mody[41];

int nn;

int com(int n, int y) {

	int ni = 1;

	for (int i = 0;i < 41;i++) {

		n2[i] = ni;

		ni *= 2;

	}

	int bunsi = 1, bunbo = 1;

	rep(0, i, y)bunsi = (bunsi * (n - i)) % mod;

	rep(0, i, y)bunbo = (bunbo * (i + 1)) % mod;

	mody[0] = bunbo;

	rep(1, i, 41) {

		bunbo = (bunbo * bunbo) % mod;

		mody[i] = bunbo;

	}

	rep(0, i, 41)nis[i] = 0;

	nn = mod - 2;

	for (int i = 40;i >= 0;i -= 1) {

		if (nn > n2[i]) {

			nis[i]++;

			nn -= n2[i];

		}

	}

	nis[0]++;

	rep(0, i, 41) {

		if (nis[i] == 1) {

			bunsi = (bunsi * mody[i]) % mod;

		}

	}

	return bunsi;

}

int gyakugen(int n, int y) {

	int ni = 1;

	for (int i = 0;i < 41;i++) {

		n2[i] = ni;

		ni *= 2;

	}

	mody[0] = y;

	rep(1, i, 41) {

		y = (y * y) % mod;

		mody[i] = y;

	}

	rep(0, i, 41)nis[i] = 0;

	nn = mod - 2;

	for (int i = 40;i >= 0;i -= 1) {

		if (nn > n2[i]) {

			nis[i]++;

			nn -= n2[i];

		}

	}

	nis[0]++;

	rep(0, i, 41) {

		if (nis[i] == 1) {

			n = (n * mody[i]) % mod;

		}

	}

	return n;

}

int yakuwa(int n) {

	int sum = 0;

	rep(1, i, sqrt(n + 1)) {

		if (n % i == 0)sum += i + n / i;

		if (i * i == n)sum -= i;

	}

	return sum;

}

int poow(int y, int n) {

	if (n == 0)return 1;

	n -= 1;

	int ni = 1;

	for (int i = 0;i < 41;i++) {

		n2[i] = ni;

		ni *= 2;

	}

	int yy = y;

	mody[0] = yy;

	rep(1, i, 41) {

		yy = (yy * yy) % mod;

		mody[i] = yy;

	}

	rep(0, i, 41)nis[i] = 0;

	nn = n;

	for (int i = 40;i >= 0;i -= 1) {

		if (nn >= n2[i]) {

			nis[i]++;

			nn -= n2[i];

		}

	}

	rep(0, i, 41) {

		if (nis[i] == 1) {

			y = (y * mody[i]) % mod;

		}

	}

	return y;

}

int minpow(int x, int y) {

	int sum = 1;

	rep(0, i, y)sum *= x;

	return sum;

}

int ketawa(int x, int sinsuu) {

	int sum = 0;

	rep(0, i, 100)sum += (x % poow(sinsuu, i + 1)) / (poow(sinsuu, i));

	return sum;

}

int sankaku(int a) {

	return a * (a + 1) / 2;

}

int sames(int a[1111111], int n) {

	int ans = 0;

	rep(0, i, n) {

		if (a[i] == a[i + 1]) {

			int j = i;

			while (a[j + 1] == a[i] && j <= n - 2)j++;

			ans += sankaku(j - i);

			i = j;

		}

	}

	return ans;

}

using Graph = vector<vector<int>>;

int oya[114514];

int depth[114514];

void dfs(const Graph& G, int v, int p, int d) {

	depth[v] = d;

	oya[v] = p;

	for (auto nv : G[v]) {

		if (nv == p) continue; // nv が親 p だったらダメ

		dfs(G, nv, v, d + 1); // d を 1 増やして子ノードへ

	}

}

/*int H=10,W=10;

char field[10][10];

char memo[10][10];

void dfs(int h, int w) {

	memo[h][w] = 'x';



	// 八方向を探索

	for (int dh = -1; dh <= 1; ++dh) {

		for (int dw = -1; dw <= 1; ++dw) {

			if(abs(0-dh)+abs(0-dw)==2)continue;

			int nh = h + dh, nw = w + dw;



			// 場外アウトしたり、0 だったりはスルー

			if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;

			if (memo[nh][nw] == 'x') continue;



			// 再帰的に探索

			dfs(nh, nw);

		}

	}

}*/

int XOR(int a, int b) {

	if (a == 0 || b == 0) {

		return a + b;

	}

	int ni = 1;

	rep(0, i, 41) {

		n2[i] = ni;

		ni *= 2;

	}

	rep(0, i, 41)nis[i] = 0;

	for (int i = 40;i >= 0;i -= 1) {

		if (a >= n2[i]) {

			nis[i]++;

			a -= n2[i];

		}

		if (b >= n2[i]) {

			nis[i]++;

			b -= n2[i];

		}

	}

	int sum = 0;

	rep(0, i, 41)sum += (nis[i] % 2 * n2[i]);

	return sum;

}

//int ma[1024577][21];

//for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=1;

struct UnionFind {

	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

 

	UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化

		for (int i = 0; i < N; i++) par[i] = i;

	}

 

	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}

		if (par[x] == x) return x;

		return par[x] = root(par[x]);

	}

 

	void unite(int x, int y) { // xとyの木を併合

		int rx = root(x); //xの根をrx

		int ry = root(y); //yの根をry

		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま

		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける

	}

 

	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す

		int rx = root(x);

		int ry = root(y);

		return rx == ry;

	}

};

int a[514514],b[514514];

signed main() {

	ic(n) ic(m)

	int ans=0;

	while(n>0){

		n-=m;

		ans++;

	}

	c(ans)

}		

// Fastest known solution for this problem: