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 {

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;
	boolean eof;

	void solve() throws IOException {
		int n = nextInt();
		int m = nextInt();
		int x = nextInt();
		
		FlowEdge[] es = new FlowEdge[m];
		int[] caps = new int[m];

		FlowGraph g = new FlowGraph(n, 0, n - 1);
		for (int i = 0; i < m; i++) {
			int v1 = nextInt() - 1;
			int v2 = nextInt() - 1;
			caps[i] = nextInt();
			es[i] = g.addDirectedEdge(v1, v2, 0);
		}
		
		double low = 0;
		double high = 1_000_000;
		
		for (int it = 0; it < 65; it++) {
			
			double mid = (low + high) * 0.5;
			for (int i = 0; i < m; i++) {
				es[i].flow = es[i].rev.flow = 0;
				es[i].cap = (int) (caps[i] / mid);
			}
			
			if (g.maxFlow() >= x) {
				low = mid;
			} else {
				high = mid;
			}
		}
		
		out.println((low + high) * 0.5 * x);
	}

	static class FlowGraph {
		int N, S, T;
		static final int INF = Integer.MAX_VALUE / 3;

		List<FlowEdge>[] g;

		private int[] dist;
		private Deque<Integer> q;
		private int[] ptr;

		public FlowGraph(int n, int s, int t) {
			N = n;
			S = s;
			T = t;
			g = new List[N];
			for (int i = 0; i < N; i++) {
				g[i] = new ArrayList<>();
			}
			dist = new int[N];
			q = new ArrayDeque<>();
			ptr = new int[N];
		}

		public FlowEdge addDirectedEdge(int v1, int v2, int cap) {
			FlowEdge e1 = new FlowEdge(v2, cap);
			FlowEdge e2 = new FlowEdge(v1, 0);
			e1.rev = e2;
			e2.rev = e1;
			g[v1].add(e1);
			g[v2].add(e2);
			return e1;
		}

		public void addUndirectedEdge(int v1, int v2, int cap) {
			FlowEdge e1 = new FlowEdge(v2, cap);
			FlowEdge e2 = new FlowEdge(v1, cap);
			e1.rev = e2;
			e2.rev = e1;
			g[v1].add(e1);
			g[v2].add(e2);
		}

		private boolean bfs() {
			Arrays.fill(dist, INF);
			dist[S] = 0;
			q.clear();
			q.add(S);

			while (!q.isEmpty()) {
				int v = q.poll();
				List<FlowEdge> adj = g[v];
				for (int i = 0; i < adj.size(); i++) {
					FlowEdge e = adj.get(i);
					if (e.flow < e.cap && dist[e.to] > dist[v] + 1) {
						dist[e.to] = dist[v] + 1;
						q.add(e.to);
					}
				}
			}
			return dist[T] != INF;
		}

		private int dfs(int v, int curFlow) {
			if (v == T)
				return curFlow;
			List<FlowEdge> adj = g[v];
			for (; ptr[v] < adj.size(); ptr[v]++) {
				FlowEdge e = adj.get(ptr[v]);
				if (dist[e.to] == dist[v] + 1 && e.flow != e.cap) {
					int go = dfs(e.to, Math.min(curFlow, e.cap - e.flow));
					if (go != 0) {
						e.flow += go;
						e.rev.flow -= go;
						return go;
					}
				}
			}
			return 0;
		}

		public int maxFlow() {
			int flow = 0;

			while (bfs()) {
				Arrays.fill(ptr, 0);
				int addFlow;
				do {
					addFlow = dfs(S, INF);
					flow += addFlow;
				} while (addFlow > 0);
			}

			return flow;
		}

	}

	static class FlowEdge {
		int to;
		int flow, cap;

		FlowEdge rev;

		public FlowEdge(int to, int cap) {
			this.to = to;
			this.cap = cap;
		}
	}

	D() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new D();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return null;
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			eof = true;
			return null;
		}
	}

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

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

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