Problem:
Score : 100 points

Problem Statement

In order to pass the entrance examination tomorrow, Taro has to study for T more hours.

Fortunately, he can leap to World B where time passes X times as fast as it does in our world (World A).

While (X \times t) hours pass in World B, t hours pass in World A.

How many hours will pass in World A while Taro studies for T hours in World B?

Constraints

All values in input are integers.

1 \leq T \leq 100

1 \leq X \leq 100

Input

Input is given from Standard Input in the following format:

T X

Output

Print the number of hours that will pass in World A.

The output will be regarded as correct when its absolute or relative error from the judge's output is at most 10^{-3}.

Sample Input 1

8 3

Sample Output 1

2.6666666667

While Taro studies for eight hours in World B where time passes three times as fast, 2.6666... hours will pass in World A.

Note that an absolute or relative error of at most 10^{-3} is allowed.

Sample Input 2

99 1

Sample Output 2

99.0000000000

Sample Input 3

1 100

Sample Output 3

0.0100000000

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _USE_MATH_DEFINES
#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <cctype>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <functional>
#include <cstdlib>
#include <iomanip>
#include <list>
#include <regex>
using namespace std;
#define rep(i, n) for (lli i = 0; i < (n); i++)
#define all(obj) (obj).begin(), (obj).end()
#define m_p make_pair
#define rond(a, b) ((a) + (b) - 1) / (b)
using lli = long long;
using vect = vector<lli>;
using vecvec = vector<vect>;
using ipair = pair<lli, lli>;
using p_q = priority_queue<lli>;
constexpr lli inf = 1e15;
constexpr lli mod = 1e9 + 7;
int dx[4] = { 1,-1,0,0 }, dy[4] = { 0,0,1,-1 };

class unionfind
{
public:
	int size;

	unionfind(int n) {
		size = n;
		for (int i = 0; i < n; i++) {
			par.push_back(i);
			rank.push_back(0);
			len.push_back(1);
		}
	};
	int find(int n) {
		if (par[n] == n)
			return n;
		else
			return par[n] = find(par[n]);
	};
	void unite(int n, int m) {
		n = find(n);
		m = find(m);
		if (n == m) return;
		size--;
		if (rank[n] < rank[m]) {
			par[n] = m;
			len[m] += len[n];
		}
		else {
			par[m] = n;
			len[n] += len[m];
		}
		if (rank[n] == rank[m])
			rank[n]++;
	};
	bool same(int n, int m) {
		return find(n) == find(m);
	};
	int length(int n) {
		return len[find(n)];
	}
private:
	vect par;
	vect rank;
	vect len;
};

class seg_tree
{
public:
	seg_tree(vect v) {
		n = 1;
		while (n < v.size()) n *= 2;
		node.resize(n * 2 - 1);
		rep(i, v.size()) node[i + n - 1] = v[i];
		for (int i = n - 2; i >= 0; i--) node[i] = node[2 * i + 1] + node[2 * i + 2];
	};
	void update(int x, lli val) {
		x += n - 1;
		node[x] = val;
		while (x > 0) {
			x = (x - 1) / 2;
			node[x] = node[2 * x + 1] + node[2 * x + 2];
		}
	};
	lli getsum(int a, int b, int k = 0, int l = 0, int r = -1) {
		if (r < 0) r = n;
		if (r <= a || b <= l) return 0;
		if (a <= l && r <= b) return node[k];
		return getsum(a, b, 2 * k + 1, l, (l + r) / 2) + getsum(a, b, 2 * k + 2, (l + r) / 2, r);
	};
private:
	int n;
	vect node;
};

class lazy_seg_tree
{
public:
	lazy_seg_tree(vect v) {
		n = 1;
		while (n < v.size()) n *= 2;
		node.resize(n * 2 - 1, inf);
		lazy.resize(n * 2 - 1, inf);
		rep(i, v.size()) node[i + n - 1] = v[i];
		for (int i = n - 2; i >= 0; i--) node[i] = min(node[2 * i + 1], node[2 * i + 2]);
	};
	void eval(int k, int l, int r) {
		if (lazy[k] != inf) {
			node[k] = min(node[k], lazy[k]);
			if (r - l > 1) {
				lazy[2 * k + 1] = min(lazy[2 * k + 1], lazy[k]);
				lazy[2 * k + 2] = min(lazy[2 * k + 1], lazy[k]);
			}
			lazy[k] = inf;
		}
	};
	void update(int a, int b, lli x, int k = 0, int l = 0, int r = -1) {
		if (r < 0) r = n;
		eval(k, l, r);
		if (b <= l || r <= a) return;
		if (a <= l && r <= b) {
			lazy[k] = min(lazy[k], x);
			eval(k, l, r);
		}
		else {
			update(a, b, x, 2 * k + 1, l, (l + r) / 2);
			update(a, b, x, 2 * k + 2, (l + r) / 2, r);
			node[k] = min(node[2 * k + 1], node[2 * k + 2]);
		}
	};
	lli getmin(int a, int b, int k = 0, int l = 0, int r = -1) {
		if (r < 0) r = n;
		if (r <= a || b <= l) return inf;
		eval(k, l, r);
		if (a <= l && r <= b) return node[k];
		return min(getmin(a, b, 2 * k + 1, l, (l + r) / 2), getmin(a, b, 2 * k + 2, (l + r) / 2, r));
	};
private:
	int n;
	vect node, lazy;
};

template<class T>
vector<pair<T, int>> rle(vector<T> vec) {
	vector<pair<T, int>> ans;
	ans.emplace_back(vec.front(), 1);
	for (int i = 1; i < vec.size(); i++) {
		if (vec[i] == ans.back().first)
			ans.back().second++;
		else
			ans.emplace_back(vec[i], 1);
	}
	return ans;
}

vect cum_sum(vect v) {
	vect ans(v.size() + 1);
	rep(i, v.size())
		ans[i + 1] = ans[i] + v[i];
	return ans;
}

vecvec cum_sum_2d(vecvec v) {
	vecvec ans(v.size() + 1, vect(v[0].size() + 1));
	rep(i, v.size())
		rep(j, v[0].size())
		ans[i + 1][j + 1] = ans[i + 1][j] + ans[i][j + 1] - ans[i][j] + v[i][j];
	return ans;
}

vector<bool> eratosthenes(int n) {
	vector<bool> ans(n + 1, 1);
	ans[1] = 0;
	for (int i = 2; i * i <= n; i++) {
		if (ans[i] == 0) continue;
		for (int j = i; i * j <= n; j++)
			ans[i * j] = 0;
	}
	return ans;
}

bool prime(int n) {
	for (int i = 2; i * i <= n; i++)
		if (n % i == 0)
			return false;
	return true;
}

vector<ipair> fact(int n) {
	vector<ipair> ans;
	auto era = eratosthenes(floor(sqrt(n)));
	for (int i = 2; i * i <= n; i++) {
		if (era[i] == 0) continue;
		if (n % i == 0) {
			ans.emplace_back(i, 1);
			n /= i;
		}
		while (n % i == 0) {
			ans.back().second++;
			n /= i;
		}
	}
	if (n != 1)
		ans.emplace_back(n, 1);
	return ans;
}

lli gcd(lli a, lli b) {
	if (a < b)
		swap(a, b);
	if (b == 0) return a;
	return gcd(b, a % b);
}

#define lcm(a, b) (a) / gcd((a), (b)) * (b)

lli s[100000], t[100000];

int main() {
	double t, x;
	cin >> t >> x;
	cout << t / x << endl;
}