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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Queue;
import java.util.StringTokenizer;

public class Maxflow {
	static long INF = (long) 1e12;

	public static void main(String[] args){
		InputReader in = new InputReader(System.in);

		OutputStream outputStream = System.out;
		PrintWriter out = new PrintWriter(outputStream);

		int n = in.nextInt();
		int m = in.nextInt();
		int x = in.nextInt();
		int source = 0;
		int sink = n - 1;

		// MaximumFlowGraph g = new MaximumFlowGraph(n, source, sink);

		for (int i = 0; i < n; i++) {
			// g.addEdge(source, i, INF);
			// g.addEdge(i + n, sink, INF);
		}
		c = new int[m];
		a = new int[m];
		b = new int[m];
		for (int j = 0; j < m; j++) {

			a[j] = in.nextInt() - 1;
			b[j] = in.nextInt() - 1;
			c[j] = in.nextInt();
			// g.addEdge(a, b, c);
		}

		// long flow = g.getFlow(source, sink);
		double ans = bs(n, m, x);
		ans = ans * x;

		// out.println(ans);
		out.printf("%.10f", ans);

		out.close();
	}

	static int[] a, b, c;

	static double bs(int n, int m, int x){
		double low = 0;
		double high = 1e6 + 5;
		double ans = 0;
		int iter = 0;
		while (true) {
			if (Math.abs(low - high) < 1e-11)
				break;
			double mid = (low + high) / 2.0;
			int source = 0;
			int sink = n - 1;
			MaximumFlowGraph g = new MaximumFlowGraph(n, source, sink);
			for (int i = 0; i < m; i++) {
				if (((int) Math.floor(c[i] / mid)) > 0)
					g.addEdge(a[i], b[i], (int) Math.floor(c[i] / mid));
			}
			long f = g.getFlow(source, sink);
			if (f >= x) {
				low = mid;
				ans = mid;
			} else {
				high = mid;
			}
			iter++;
			if (iter > 100)
				break;
		}
		return ans;

	}

	static class MaximumFlowGraph {

		final long INF = (long) 1e15;

		public class Edge {
			public int from;
			public int to;
			public long flow;
			public long cap;

			Edge(int from, int to, long cap) {
				this.from = from;
				this.to = to;
				this.cap = cap;
				this.flow = 0;
			}

			boolean isOpen(){
				return cap > flow;
			}

			long getLeft(){
				return cap - flow;
			}
		}

		public ArrayList<Edge> e = new ArrayList<>();
		long[] dist;
		int[] parEdge;
		int[] first;
		ArrayList<Integer>[] g;
		int source;
		int sink;

		public void setzeroflow(){
			for (int i = 0; i < e.size(); i++) {
				Edge ee = e.get(i);
				ee.flow = 0;
			}
		}

		public MaximumFlowGraph(int size, int source, int sink) {
			this.source = source;
			this.sink = sink;
			dist = new long[size];
			parEdge = new int[size];
			first = new int[size];
			g = new ArrayList[size];
			for (int i = 0; i < size; i++)
				g[i] = new ArrayList<>();
		}

		public void addEdge(int from, int to, long cap){
			int size = e.size();
			e.add(new Edge(from, to, cap));
			e.add(new Edge(to, from, 0));
			g[from].add(size);
			g[to].add(size + 1);
		}

		private boolean bfs(){
			Arrays.fill(dist, INF);
			dist[source] = 0;
			Queue<Integer> q = new ArrayDeque<>();
			q.add(source);
			while (!q.isEmpty()) {
				int v = q.poll();
				for (int j : g[v]) {
					Edge curr = e.get(j);
					int to = curr.to;

					if (curr.isOpen() && dist[to] > dist[v] + 1) {
						dist[to] = dist[v] + 1;
						q.add(to);
						if (to == sink)
							return true;

					}

				}
			}
			return dist[sink] != INF;
		}

		private boolean dfs(int v, int pe){
			parEdge[v] = pe;
			if (v == sink)
				return true;
			if (dist[v] >= dist[sink])
				return false;
			for (; first[v] < g[v].size(); first[v]++) {
				Edge curr = e.get(g[v].get(first[v]));
				if (curr.isOpen() && dist[curr.to] == dist[v] + 1 && dfs(curr.to, g[v].get(first[v])))
					return true;
			}
			return false;
		}

		long pushFlow(){
			long add = INF;
			int v = sink;
			while (v != source) {
				Edge curr = e.get(parEdge[v]);
				add = Math.min(add, curr.getLeft());
				v = curr.from;
			}
			v = sink;
			while (v != source) {
				int i = parEdge[v];
				e.get(i).flow += add;
				e.get(i ^ 1).flow -= add;
				v = e.get(i).from;
			}
			return add;
		}

		public long getFlow(int source, int sink){

			this.source = source;
			this.sink = sink;

			long flow = 0;
			while (bfs()) {
				Arrays.fill(first, 0);
				while (dfs(source, -1))
					flow += pushFlow();
			}
			return flow;
		}

	}

	public static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream inputstream) {
			reader = new BufferedReader(new InputStreamReader(inputstream));
			tokenizer = null;
		}

		public String nextLine(){
			String fullLine = null;
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					fullLine = reader.readLine();
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
				return fullLine;
			}
			return fullLine;
		}

		public String next(){
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

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

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