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.util.*;
import java.io.*;

public class d {

	public static ArrayList[] graph;
	public static int n;
	public static int numBears;

	public static void main(String[] args) throws Exception {

		BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer tok = new StringTokenizer(stdin.readLine());
		n = Integer.parseInt(tok.nextToken());
		int e = Integer.parseInt(tok.nextToken());
		numBears = Integer.parseInt(tok.nextToken());

		graph = new ArrayList[n];
		for (int i=0; i<n; i++) graph[i] = new ArrayList<Edge>();

		int cut = 0;
		for (int i=0; i<e; i++) {
			tok = new StringTokenizer(stdin.readLine());
			int v1 = Integer.parseInt(tok.nextToken())-1;
			int v2 = Integer.parseInt(tok.nextToken())-1;
			int w = Integer.parseInt(tok.nextToken());
			graph[v1].add(new Edge(w, v2));
			if (v1 == 0) cut += w;
		}

		double low = 0, high = cut;

		for (int loop=0; loop<50; loop++) {

			double mid = (low+high)/2;

			if (possible(mid))
				low = mid;
			else
				high = mid;
		}

		System.out.println(low);

	}

	public static boolean possible(double x) {

		int[][] adj = new int[n+1][n+1];
		adj[0][1] = numBears;
		double perBear = x/numBears;

		for (int i=0; i<graph.length; i++) {
			for (int j=0; j<graph[i].size(); j++) {
				int to = ((ArrayList<Edge>)graph[i]).get(j).dest;
				int neww = (int) ( ((ArrayList<Edge>)graph[i]).get(j).capacity/perBear + 1e-9);
				adj[i+1][to+1] = neww;
			}
		}

		netflow myFlow = new netflow(adj, 0, n);
		int res = myFlow.getMaxFlow();
		return res == numBears;

	}
}

/*** My network flow code ***/
class Edge {

	public int dest;
	public int capacity;
	public int flow;

	public Edge(int cap, int d) {
		capacity = cap;
		flow = 0;
		dest = d;
	}

	public int maxPushForward() {
		return capacity - flow;
	}

	public int maxPushBackward() {
		return flow;
	}

	public boolean pushForward(int moreflow) {

		// We can't push through this much flow.
		if (flow+moreflow > capacity)
			return false;

		// Push through.
		flow += moreflow;
		return true;
	}

	public boolean pushBack(int lessflow) {

		// Not enough to push back on.
		if (flow < lessflow)
			return false;

		flow -= lessflow;
		return true;
	}
}

class direction {

	public int prev;
	public boolean forward;

	public direction(int node, boolean dir) {
		prev = node;
		forward = dir;
	}

	public String toString() {
		if (forward)
			return "" + prev + "->";
		else
			return "" + prev + "<-";
	}
}

class netflow {

	private Edge[][] adjMat;
	private int source;
	private int dest;
	private HashMap[] lookup;
	private LinkedList[] backEdgeLookup;

	// All positive entries in flows should represent valid flows
	// between vertices. All other entries must be 0 or negative.
	public netflow(int[][] flows, int start, int end) {

		source = start;
		dest = end;
		adjMat = new Edge[flows.length][];
		lookup = new HashMap[flows.length];

		backEdgeLookup = new LinkedList[flows.length];
		for (int i=0; i<flows.length; i++)
			backEdgeLookup[i] = new LinkedList<Integer>();

		for (int i=0; i<flows.length; i++) {

			lookup[i] = new HashMap<Integer,Integer>();

			int numEdges = 0;
			for (int j=0; j<flows[i].length; j++)
				if (flows[i][j] > 0)
					numEdges++;

			adjMat[i] = new Edge[numEdges];

			int cnt = 0;
			for (int j=0; j<flows[i].length; j++) {

				// Fill in this flow.
				if (flows[i][j] > 0) {
					lookup[i].put(j, cnt);
					adjMat[i][cnt++] = new Edge(flows[i][j], j);
					backEdgeLookup[j].offer(i);
				}
			}
		}
	}

	public ArrayList<direction> findAugmentingPath() {

		// This will store the previous node visited in the BFS.
		direction[] prev = new direction[adjMat.length];
		boolean[] inQueue = new boolean[adjMat.length];
		for (int i=0; i<inQueue.length; i++)
			inQueue[i] = false;

		// The source has no previous node.
		prev[source] = new direction(-1, true);

		LinkedList<Integer> bfs_queue = new LinkedList<Integer>();
		bfs_queue.offer(new Integer(source));
		inQueue[source] = true;

		// Our BFS will go until we clear out the queue.
		while (bfs_queue.size() > 0) {

			// Add all the new neighbors of the current node.
			Integer next = bfs_queue.poll();


			// Find all neighbors and add into the queue. These are forward edges.
			for (int i=0; i<adjMat[next].length; i++) {

				int item = adjMat[next][i].dest;
				if (!inQueue[item] && adjMat[next][i].maxPushForward() > 0) {
					bfs_queue.offer(item);
					inQueue[item] = true;
					prev[item] = new direction(next, true);
				}
			}

			/*** We really want i to next. ***/
			// Now look for back edges.
			for (int i=0; i<backEdgeLookup[next].size(); i++) {

				int item = (Integer)backEdgeLookup[next].pollFirst();
				if (!inQueue[item] && lookup[item].containsKey(next) && adjMat[item][(Integer)(lookup[item].get(next))].maxPushBackward() > 0) {
					bfs_queue.offer(item);
					inQueue[item] = true;
					prev[item] = new direction(next, false);
				}
				backEdgeLookup[next].offer(item);
			}
		}

		// No augmenting path found.
		if (!inQueue[dest])
			return null;

		ArrayList<direction> path = new ArrayList<direction>();

		direction place = prev[dest];

		direction dummy = new direction(dest, true);
		path.add(dummy);

		// Build the path backwards.
		while (place.prev != -1) {
			path.add(place);
			place = prev[place.prev];
		}

		// Reverse it now.
		Collections.reverse(path);

		return path;
	}

	// Run the Max Flow Algorithm here.
	public int getMaxFlow() {

		int flow = 0;

		ArrayList<direction> nextpath = findAugmentingPath();

		// Loop until there are no more augmenting paths.
		while (nextpath != null) {

			// Check what the best flow through this path is.
			int this_flow = Integer.MAX_VALUE;
			for (int i=0; i<nextpath.size()-1; i++) {

				if (nextpath.get(i).forward) {
					this_flow = Math.min(this_flow, adjMat[nextpath.get(i).prev][(Integer)lookup[nextpath.get(i).prev].get(nextpath.get(i+1).prev)].maxPushForward());
				}
				else {
					this_flow = Math.min(this_flow, adjMat[nextpath.get(i+1).prev][(Integer)lookup[nextpath.get(i+1).prev].get(nextpath.get(i).prev)].maxPushBackward());
				}
			}

			// Now, put this flow through.
			for (int i=0; i<nextpath.size()-1; i++) {

				if (nextpath.get(i).forward) {
					adjMat[nextpath.get(i).prev][(Integer)lookup[nextpath.get(i).prev].get(nextpath.get(i+1).prev)].pushForward(this_flow);
				}
				else {
					adjMat[nextpath.get(i+1).prev][(Integer)lookup[nextpath.get(i+1).prev].get(nextpath.get(i).prev)].pushBack(this_flow);
				}
			}

			// Add this flow in and then get the next path.
			flow += this_flow;
			nextpath = findAugmentingPath();
		}

		return flow;
	}

}