Problem: Niwel is a little golden bear. As everyone knows, bears live in forests, but Niwel got tired of seeing all the trees so he decided to move to the city.

In the city, Niwel took on a job managing bears to deliver goods. The city that he lives in can be represented as a directed graph with n nodes and m edges. Each edge has a weight capacity. A delivery consists of a bear carrying weights with their bear hands on a simple path from node 1 to node n. The total weight that travels across a particular edge must not exceed the weight capacity of that edge.

Niwel has exactly x bears. In the interest of fairness, no bear can rest, and the weight that each bear carries must be exactly the same. However, each bear may take different paths if they like.

Niwel would like to determine, what is the maximum amount of weight he can deliver (it's the sum of weights carried by bears). Find the maximum weight.

Input

The first line contains three integers n, m and x (2 ≤ n ≤ 50, 1 ≤ m ≤ 500, 1 ≤ x ≤ 100 000) — the number of nodes, the number of directed edges and the number of bears, respectively.

Each of the following m lines contains three integers ai, bi and ci (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ ci ≤ 1 000 000). This represents a directed edge from node ai to bi with weight capacity ci. There are no self loops and no multiple edges from one city to the other city. More formally, for each i and j that i ≠ j it's guaranteed that ai ≠ aj or bi ≠ bj. It is also guaranteed that there is at least one path from node 1 to node n.

Output

Print one real value on a single line — the maximum amount of weight Niwel can deliver if he uses exactly x bears. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct if <image>.

Examples

Input

4 4 3
1 2 2
2 4 1
1 3 1
3 4 2


Output

1.5000000000


Input

5 11 23
1 2 3
2 3 4
3 4 5
4 5 6
1 3 4
2 4 5
3 5 6
1 4 2
2 5 3
1 5 2
3 2 30


Output

10.2222222222

Note

In the first sample, Niwel has three bears. Two bears can choose the path <image>, while one bear can choose the path <image>. Even though the bear that goes on the path <image> can carry one unit of weight, in the interest of fairness, he is restricted to carry 0.5 units of weight. Thus, the total weight is 1.5 units overall. Note that even though Niwel can deliver more weight with just 2 bears, he must use exactly 3 bears on this day.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);

		int n = sc.nextInt();
		int m = sc.nextInt();
		int x = sc.nextInt();
		int[] a = new int[n];
		V = n;
		s = 0;
		t = n - 1;
		res = new double[V][V];
		adjList = new ArrayList[V];
		for (int i = 0; i < V; i++) {
			adjList[i] = new ArrayList<>();
		}
		for (int i = 0; i < m; i++) {
			int u = sc.nextInt() - 1;
			int v = sc.nextInt() - 1;
			int w = sc.nextInt();
			adjList[u].add(v);
			res[u][v] = w;
		}

		double s = 0;
		double e = 1e9;
		double ans = 0;
//		int xx = dinic(0.625);
//		System.out.println(xx );
		for (int i = 0; i < 100; i++) {
			double mid = (s + e) / 2;
			int sum = dinic(mid);
//			System.out.println(sum +" "+  mid);
			if (sum >= x) {
				ans = mid * x;
				s = mid;
			} else
				e = mid;
		}
		pw.println(ans);

		pw.flush();
	}

	static final int INF = (int) 1e9;
	static int V, s, t;
	static double res[][]; // input
	static double test[][];
	static ArrayList<Integer>[] adjList; // input
	static int[] ptr, dist;

	static int dinic(double mid) // O(V^2E)
	{
//		System.out.println(Arrays.toString(res[0]));
		test=new double[V][V];
		for (int i = 0; i < test.length; i++) {
			test[i] = res[i].clone();
		}
		int mf = 0;
		while (bfs(mid)) {
			ptr = new int[V];
			double f;
			while ((f = dfs(s, mid)) != 0)
				mf++;
		}
		return mf;
	}

	static boolean bfs(double mid) {
		dist = new int[V];
		Arrays.fill(dist, -1);
		dist[s] = 0;
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(s);
		while (!q.isEmpty()) {
			int u = q.remove();
			if (u == t)
				return true;
			for (int v : adjList[u])
				if (dist[v] == -1 && (Math.abs(test[u][v] - mid) <= 1e-7 || test[u][v] >= mid)) {
					dist[v] = dist[u] + 1;
					q.add(v);
				}
		}
		return false;
	}

	static ArrayList<ArrayList<Integer>> ans;

	static double dfs(int u, double mid) {
		if (u == t) {
			return 1;
		}
		for (int i = ptr[u]; i < adjList[u].size(); i = ++ptr[u]) {
			int v = adjList[u].get(i);
			if (dist[v] == dist[u] + 1 && (Math.abs(test[u][v] - mid) <= 1e-7 || test[u][v] > mid)) {
				double f = dfs(v, mid);
				if (f > 0) {
					test[u][v] -= mid;
					test[v][u] += mid;
					return f;
				}
			}
		}
		return 0;

	}

	static int[] isComposite;

	static void sieve(int N) // O(N log log N)
	{
		isComposite = new int[N + 1];
		isComposite[0] = isComposite[1] = 1; // 0 indicates a prime number
//		primes = new ArrayList<Integer>();

		for (int i = 2; i <= N; ++i) // can loop till i*i <= N if primes array is not needed O(N log log sqrt(N))
			if (isComposite[i] == 0) // can loop in 2 and odd integers for slightly better performance
			{
//				primes.add(i);
				if (1l * i * i <= N)
					for (int j = i * i; j <= N; j += i) // j = i * 2 will not affect performance too much, may alter in
														// modified sieve
						isComposite[j] = 1;
			}
	}

	static class pair implements Comparable<pair> {
		int x, y;
		int w;

		pair(int s, int d) {
			x = s;
			y = d;
		}

		pair(int s, int d, int c) {
			x = s;
			y = d;
			w = c;
		}

		@Override
		public int compareTo(pair p) {
			return (x == p.x && y == p.y) ? 0 : 1;
		}

		@Override
		public String toString() {
			return x + " " + y;

		}

	}

	static long mod(long ans, int mod) {
		return (ans % mod + mod) % mod;
	}

	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}

	public static int log(int n, int base) {
		int ans = 0;
		while (n + 1 > base) {
			ans++;
			n /= base;
		}
		return ans;
	}

	static long pow(long b, long e) {
		long ans = 1;
		while (e > 0) {
			if ((e & 1) == 1)
				ans = ((ans * 1l * b));
			e >>= 1;
			{

			}
			b = ((b * 1l * b));
		}
		return ans;
	}

	static int powmod(int b, long e, int mod) {
		int ans = 1;
		b %= mod;
		while (e > 0) {
			if ((e & 1) == 1)
				ans = (int) ((ans * 1l * b) % mod);
			e >>= 1;
			b = (int) ((b * 1l * b) % mod);
		}
		return ans;
	}

	static int ceil(int a, int b) {
		int ans = a / b;
		return a % b == 0 ? ans : ans + 1;
	}

	static long ceil(long a, long b) {
		long ans = a / b;
		return a % b == 0 ? ans : ans + 1;
	}

	static HashMap<Integer, Integer> compress(int a[]) {
		TreeSet<Integer> ts = new TreeSet<>();
		HashMap<Integer, Integer> hm = new HashMap<>();
		for (int x : a)
			ts.add(x);
		for (int x : ts) {
			hm.put(x, hm.size() + 1);
		}
		return hm;
	}

	// Returns nCr % p
	static int C[];

	static int nCrModp(int n, int r, int p) {
		if (r > n - r)
			r = n - r;
		if (C[r] != 0)
			return C[r];
		// The array C is going to store last
		// row of pascal triangle at the end.
		// And last entry of last row is nCr

		C[0] = 1; // Top row of Pascal Triangle

		// One by constructs remaining rows of Pascal
		// Triangle from top to bottom
		for (int i = 1; i <= n; i++) {

			// Fill entries of current row using previous
			// row values
			for (int j = Math.min(i, r); j > 0; j--)

				// nCj = (n-1)Cj + (n-1)C(j-1);
				C[j] = (C[j] + C[j - 1]) % p;
		}
		return C[r];
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public int[] intArr(int n) throws IOException {
			int a[] = new int[n];
			for (int i = 0; i < a.length; i++) {
				a[i] = nextInt();
			}
			return a;
		}

		public long[] longArr(int n) throws IOException {
			long a[] = new long[n];
			for (int i = 0; i < a.length; i++) {
				a[i] = nextLong();
			}
			return a;
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

	}

	public static void shuffle(Integer[] x2) {
		int n = x2.length;
		for (int i = 0; i < n; i++) {
			int r = i + (int) (Math.random() * (n - i));
			int tmp = x2[i];
			x2[i] = x2[r];
			x2[r] = tmp;
		}
	}

}