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 a_{i}, b_{i} and c_{i} (1 ≤ a_{i}, b_{i} ≤ n, a_{i} ≠ b_{i}, 1 ≤ c_{i} ≤ 1 000 000). This represents a directed edge from node a_{i} to b_{i} with weight capacity c_{i}. 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 a_{i} ≠ a_{j} or b_{i} ≠ b_{j}. 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 $\frac{|a - b|}{\operatorname{max}(1, b)} \leq 10^{-6}$.


-----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 $1 \rightarrow 3 \rightarrow 4$, while one bear can choose the path $1 \rightarrow 2 \rightarrow 4$. Even though the bear that goes on the path $1 \rightarrow 2 \rightarrow 4$ 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.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'from collections import defaultdict, deque\n \n \ndef bfs(graph, inicio, destino, parent):\n    parent.clear()\n    queue = deque()\n    queue.append([inicio, float("Inf")])\n    parent[inicio] = -2\n    while (len(queue)):\n        current, flow = queue.popleft()\n        for i in graph[current]:\n            if parent[i] == -1 and graph[current][i] > 0:\n                parent[i] = current\n                flow = min(flow, graph[current][i])\n                if i == destino:\n                    return flow\n                queue.append((i, flow))\n    return 0\n \n \ndef maxflow(graph, inicio, destino):\n    flow = 0\n    parent = defaultdict(lambda: -1)\n    while True:\n        t = bfs(graph, inicio, destino, parent)\n        if t:\n            flow += t\n            current = destino\n            while current != inicio:\n                prev = parent[current]\n                graph[prev][current] -= t\n                graph[current][prev] += t\n                current = prev\n        else:\n            break\n    return flow\n \n \nn, m, x = [int(i) for i in input().split()]\ngraph = defaultdict(lambda: defaultdict(lambda: 0))\nfor _ in range(m):\n    t = [int(i) for i in input().split()]\n    graph[t[0]][t[1]] = t[2]\n \n \ndef check(k):\n    meh = defaultdict(lambda: defaultdict(lambda: 0))\n    for i in graph:\n        for j in graph[i]:\n            ww = graph[i][j] // k\n            meh[i][j] = ww\n    flow = maxflow(meh, 1, n)\n    return flow\n \n \nlo = 1 / x\nhi = check(1)\n \nfor _ in range(70):\n    mid = round((hi + lo) / 2,9)\n    if check(mid)>=x:\n        lo = round(mid,9)\n    else:\n        hi = mid\nprint(format(lo * x, \'.9f\'))'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14221