Problem:
Score : 500 points

Problem Statement

There is a directed graph with N vertices numbered 1 to N and M edges.
The i-th edge is directed from Vertex A_i to Vertex B_i, and there are C_i coins placed along that edge.
Additionally, there is a button on Vertex N.

We will play a game on this graph.
You start the game on Vertex 1 with zero coins, and head for Vertex N by traversing the edges while collecting coins.
It takes one minute to traverse an edge, and you can collect the coins placed along the edge each time you traverse it.
As usual in games, even if you traverse an edge once and collect the coins, the same number of coins will reappear next time you traverse that edge, which you can collect again.

When you reach Vertex N, you can end the game by pressing the button. (You can also choose to leave Vertex N without pressing the button and continue traveling.)
However, when you end the game, you will be asked to pay T \times P coins, where T is the number of minutes elapsed since the start of the game. If you have less than T \times P coins, you will have to pay all of your coins instead.

Your score will be the number of coins you have after this payment.
Determine if there exists a maximum value of the score that can be obtained. If the answer is yes, find that maximum value.

Constraints

2 \leq N \leq 2500

1 \leq M \leq 5000

1 \leq A_i, B_i \leq N

1 \leq C_i \leq 10^5

0 \leq P \leq 10^5

All values in input are integers.

Vertex N can be reached from Vertex 1.

Input

Input is given from Standard Input in the following format:

N M P
A_1 B_1 C_1
:
A_M B_M C_M

Output

If there exists a maximum value of the score that can be obtained, print that maximum value; otherwise, print -1.

Sample Input 1

3 3 10
1 2 20
2 3 30
1 3 45

Sample Output 1

35

There are two ways to travel from Vertex 1 to Vertex 3:

Vertex 1 \rightarrow 2 \rightarrow 3: You collect 20 + 30 = 50 coins on the way. After two minutes from the start of the game, you press the button, pay 2 \times 10 = 20 coins, and you have 50 - 20 = 30 coins left.

Vertex 1 \rightarrow 2: You collect 45 coins on the way. After one minute from the start of the game, you press the button, pay 1 \times 10 = 10 coins, and you have 45 - 10 = 35 coins left.

Thus, the maximum score that can be obtained is 35.

Sample Input 2

2 2 10
1 2 100
2 2 100

Sample Output 2

-1

The edge extending from Vertex 1 takes you to Vertex 2. If you then traverse the edge extending from Vertex 2 to itself t times and press the button, your score will be 90 + 90t. Thus, you can infinitely increase your score, which means there is no maximum value of the score that can be obtained.

Sample Input 3

4 5 10
1 2 1
1 4 1
3 4 1
2 2 100
3 3 100

Sample Output 3

0

There is no way to travel from Vertex 1 to Vertex 4 other than traversing the edge leading from Vertex 1 to Vertex 4 directly. You will pick up one coin along this edge, but after being asked to paying 10 coins, your score will be 0.

Note that you can collect an infinite number of coins if you traverse the edge leading from Vertex 1 to Vertex 2, but this is pointless since you can no longer reach Vertex 4 and end the game.

Solution in C# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Globalization;
using System.IO;
using System.Linq;

using E = System.Linq.Enumerable;

using CostType = System.Int64;

internal partial class Solver {
    public void Run() {
        var n = ni();
        var m = ni();
        var p = ni();
        var g = new WeightedDirectedGraph(n);
        for (int i = 0; i < m; i++) {
            var a = ni() - 1;
            var b = ni() - 1;
            var c = ni();
            g.AddEdge(a, b, p - c);
        }
        var ans = g.BellmanFord(0)[n - 1];
        if (ans == CostType.MinValue) {
            ans = -1;
        } else {
            ans = Math.Max(0, -ans);
        }
        cout.WriteLine(ans);
    }
}


public class WeightedDirectedGraph {
    public struct Edge {
        public int From { get; set; }
        public int To { get; set; }
        public CostType Cost { get; set; }
    }
    private readonly List<Edge>[] _adj;
    public WeightedDirectedGraph(int n) {
        _adj = Enumerable.Range(0, n).Select(_ => new List<Edge>()).ToArray();
    }

    public void AddEdge(int from, int to, CostType cost) {
        _adj[from].Add(new Edge { From = from, To = to, Cost = cost });
    }

    public IEnumerable<Edge> this[int index] {
        get { return _adj[index]; }
    }

    public int VertexCount { get { return _adj.Length; } }
}

public static partial class ShortestPathExtension {
    public static ShorestPathInfo BellmanFord(this WeightedDirectedGraph graph, int start) {
        var n = graph.VertexCount;
        var dist = new CostType[n];
        var prev = Enumerable.Repeat(-1, n).ToArray();
        var inf = CostType.MaxValue;
        var negativeInf = CostType.MinValue;
        for (int i = 0; i < n; i++) dist[i] = inf;
        dist[start] = 0;
        for (int i = 0; i <= n + 10; i++) {
            bool updated = false;
            for (int j = 0; j < n; j++) {
                foreach (var e in graph[j]) {
                    var newCost = dist[e.From];
                    if (newCost != inf) {
                        if (newCost != negativeInf) newCost += e.Cost;
                        if (dist[e.To] > newCost) {
                            prev[e.To] = e.From;
                            dist[e.To] = i >= n ? negativeInf : newCost;
                            updated = true;
                        }
                    }
                }
            }
            if (!updated) break;
        }
        var queue = new Queue<int>();
        for (int i = 0; i < n; i++) {
            if (dist[i] == negativeInf) {
                queue.Enqueue(i);
            }
        }
        while (queue.Count > 0) {
            var now = queue.Dequeue();
            if (dist[now] != negativeInf) continue;
            dist[now] = negativeInf;
            foreach (var e in graph[now]) {
                if (dist[e.To] != negativeInf) {
                    queue.Enqueue(e.To);
                }
            }
        }

        return new ShorestPathInfo(dist, prev);
    }


    public static ShorestPathInfo SPFA(this WeightedDirectedGraph graph, int start) {
        var n = graph.VertexCount;
        var inf = CostType.MaxValue;
        var dist = Enumerable.Repeat(inf, n).ToArray();
        var prev = Enumerable.Repeat(-1, n).ToArray();
        var inQueue = new bool[n];
        var visitedNum = new int[n];
        var qu = new Queue<int>();
        dist[start] = 0;
        inQueue[start] = true;
        qu.Enqueue(start);
        while (qu.Count > 0) {
            var pos = qu.Dequeue();
            inQueue[pos] = false;
            foreach (var edge in graph[pos]) {
                if (dist[pos] != inf && dist[edge.To] > dist[pos] + edge.Cost) {
                    dist[edge.To] = dist[pos] + edge.Cost;
                    prev[edge.To] = pos;
                    if (!inQueue[edge.To]) {
                        if (++visitedNum[edge.To] < n) {
                            inQueue[edge.To] = true;
                            qu.Enqueue(edge.To);
                        }
                    }
                }
            }
        }
        return new ShorestPathInfo(dist, prev);
    }
}

public class ShorestPathInfo {
    private readonly CostType[] _dist;
    private readonly int[] _prev;
    public ShorestPathInfo(CostType[] dist, int[] prev) {
        _dist = dist;
        _prev = prev;
    }
    public CostType this[int index] {
        get { return _dist[index]; }
    }
    public IEnumerable<int> GetPath(int to) {
        int current = to;
        var path = new List<int>();
        do {
            path.Add(current);
            current = _prev[current];
        } while (current >= 0);
        path.Reverse();
        return path;
    }
}

// PREWRITEN CODE BEGINS FROM HERE
internal partial class Solver : Scanner {
    public static void Main(string[] args) {
#if LOCAL
        new Solver(Console.In, Console.Out).Run();
#else
        Console.SetOut(new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false });
        new Solver(Console.In, Console.Out).Run();
        Console.Out.Flush();
#endif
    }

#pragma warning disable IDE0052
    private readonly TextReader cin;
    private readonly TextWriter cout;
#pragma warning restore IDE0052

    public Solver(TextReader reader, TextWriter writer)
        : base(reader) {
        cin = reader;
        cout = writer;
    }
    public Solver(string input, TextWriter writer)
        : this(new StringReader(input), writer) {
    }

#pragma warning disable IDE1006
#pragma warning disable IDE0051
    private int ni() { return NextInt(); }
    private int[] ni(int n) { return NextIntArray(n); }
    private long nl() { return NextLong(); }
    private long[] nl(int n) { return NextLongArray(n); }
    private double nd() { return NextDouble(); }
    private double[] nd(int n) { return NextDoubleArray(n); }
    private string ns() { return Next(); }
    private string[] ns(int n) { return NextArray(n); }
#pragma warning restore IDE1006
#pragma warning restore IDE0051
}

internal static class LinqPadExtension {
    public static T Dump<T>(this T obj) {
#if LOCAL
        return LINQPad.Extensions.Dump(obj);
#else
        return obj;
#endif
    }
}

public class Scanner {
    private readonly TextReader Reader;
    private readonly Queue<string> TokenQueue = new Queue<string>();
    private readonly CultureInfo ci = CultureInfo.InvariantCulture;

    public Scanner()
        : this(Console.In) {
    }

    public Scanner(TextReader reader) {
        Reader = reader;
    }

    public int NextInt() { return int.Parse(Next(), ci); }
    public long NextLong() { return long.Parse(Next(), ci); }
    public double NextDouble() { return double.Parse(Next(), ci); }
    public string[] NextArray(int size) {
        string[] array = new string[size];
        for (int i = 0; i < size; i++) {
            array[i] = Next();
        }

        return array;
    }
    public int[] NextIntArray(int size) {
        int[] array = new int[size];
        for (int i = 0; i < size; i++) {
            array[i] = NextInt();
        }

        return array;
    }

    public long[] NextLongArray(int size) {
        long[] array = new long[size];
        for (int i = 0; i < size; i++) {
            array[i] = NextLong();
        }

        return array;
    }

    public double[] NextDoubleArray(int size) {
        double[] array = new double[size];
        for (int i = 0; i < size; i++) {
            array[i] = NextDouble();
        }

        return array;
    }

    public string Next() {
        if (TokenQueue.Count == 0) {
            if (!StockTokens()) {
                throw new InvalidOperationException();
            }
        }
        return TokenQueue.Dequeue();
    }

    public bool HasNext() {
        if (TokenQueue.Count > 0) {
            return true;
        }

        return StockTokens();
    }

    private static readonly char[] _separator = new[] { ' ', '\t' };
    private bool StockTokens() {
        while (true) {
            string line = Reader.ReadLine();
            if (line == null) {
                return false;
            }

            string[] tokens = line.Split(_separator, StringSplitOptions.RemoveEmptyEntries);
            if (tokens.Length == 0) {
                continue;
            }

            foreach (string token in tokens) {
                TokenQueue.Enqueue(token);
            }

            return true;
        }
    }
}