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 d {
public static void main(String[] args) throws IOException {
	input.init(System.in);
	PrintWriter out = new PrintWriter(System.out);
	int n = input.nextInt(), m = input.nextInt(), x = input.nextInt();
	ArrayList<Edge>[] g = new ArrayList[n];
	for(int i = 0; i<n; i++) g[i] = new ArrayList<Edge>();
	for(int i = 0; i<m; i++)
	{
		int a = input.nextInt()-1, b = input.nextInt()-1, c = input.nextInt();
		g[a].add(new Edge(b, c));
	}
	double lo = 0, hi = 1e6+1;
	for(int iter = 0; iter < 60; iter++)
	{
		double mid = (lo + hi)/2;
		TidalFlow tf = new TidalFlow(n);
		tf.add(tf.s, 0, x);
		tf.add(n-1, tf.t, x);
		for(int i = 0; i<n; i++)
			for(Edge e : g[i])
			{
				tf.add(i, e.to, (int)(e.w / mid));
			}
		if(tf.getFlow() == x) lo = mid;
		else hi = mid;
	}
	out.println(x * lo);
	out.close();
}
static class Edge
{
	int to, w;
	public Edge(int tt, int ww)
	{
		to = tt; w = ww;
	}
}
static class TidalFlow {
	ArrayDeque<Edge> stk = new ArrayDeque<Edge>();
	int N, s, t, oo = 987654321, fptr, bptr;
	ArrayList<Edge>[] adj;
	int[] q, dist, pool;

	@SuppressWarnings("unchecked")
	TidalFlow(int NN) {
		N=(t=(s=NN)+1)+1;
		adj = new ArrayList[N];
		for(int i = 0; i < N; adj[i++] = new ArrayList<Edge>());
		dist = new int[N];
		pool = new int[N];
		q = new int[N];
	}
	void add(int i, int j, int cap) {
		Edge fwd = new Edge(i, j, cap, 0);
		Edge rev = new Edge(j, i, 0, 0);
		adj[i].add(rev.rev=fwd);
		adj[j].add(fwd.rev=rev);
	}
	int augment() {
		Arrays.fill(dist, Integer.MAX_VALUE);
		pool[t] = dist[s] = fptr = bptr = 0;
		pool[q[bptr++] = s] = oo;
		while(bptr > fptr && q[fptr] != t)
			for(Edge e : adj[q[fptr++]]) {
				if(dist[e.i] < dist[e.j])
					pool[e.j] += e.carry = Math.min(e.cap - e.flow, pool[e.i]);
				if(dist[e.i] + 1 < dist[e.j] && e.cap > e.flow)
					dist[q[bptr++] = e.j] = dist[e.i] + 1;
			}
		if(pool[t] == 0) return 0;
		Arrays.fill(pool, fptr = bptr = 0);
		pool[q[bptr++] = t] = oo;
		while(bptr > fptr) 
			for(Edge e : adj[q[fptr++]]) {
				if(pool[e.i] == 0) break;
				int f = e.rev.carry = Math.min(pool[e.i], e.rev.carry);
				if(dist[e.i] > dist[e.j] && f != 0) {
					if(pool[e.j] == 0) q[bptr++] = e.j;
					pool[e.i] -= f;
					pool[e.j] += f;
					stk.push(e.rev);
				}
			}
		int res = pool[s];
		Arrays.fill(pool, 0);
		pool[s] = res;
		while(stk.size() > 0) {
			Edge e = stk.pop();
			int f = Math.min(e.carry, pool[e.i]);
			pool[e.i] -= f;
			pool[e.j] += f;
			e.flow += f;
			e.rev.flow -= f;
		}
		return res;
	}
	int getFlow() {
		int res = 0, f = 1;
		while(f != 0)
			res += f = augment();
		return res;
	}
	class Edge {
		int i, j, cap, flow, carry;
		Edge rev;
		Edge(int ii, int jj, int cc, int ff) {
			i=ii; j=jj; cap=cc; flow=ff;
		}
	}
}
public static class input {
	static BufferedReader reader;
	static StringTokenizer tokenizer;

	static void init(InputStream input) {
		reader = new BufferedReader(new InputStreamReader(input));
		tokenizer = new StringTokenizer("");
	}

	static String next() throws IOException {
		while (!tokenizer.hasMoreTokens())
			tokenizer = new StringTokenizer(reader.readLine());
		return tokenizer.nextToken();
	}

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

	static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

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