Problem:
Score : 900 points

Problem Statement

Takahashi has an N \times N grid. The square at the i-th row and the j-th column of the grid is denoted by (i,j).
Particularly, the top-left square of the grid is (1,1), and the bottom-right square is (N,N).

An integer, 0 or 1, is written on M of the squares in the Takahashi's grid.
Three integers a_i,b_i and c_i describe the i-th of those squares with integers written on them: the integer c_i is written on the square (a_i,b_i).

Takahashi decides to write an integer, 0 or 1, on each of the remaining squares so that the condition below is satisfied.
Find the number of such ways to write integers, modulo 998244353.

For all 1\leq i < j\leq N, there are even number of 1s in the square region whose top-left square is (i,i) and whose bottom-right square is (j,j).

Constraints

2 \leq N \leq 10^5

0 \leq M \leq min(5 \times 10^4,N^2)

1 \leq a_i,b_i \leq N(1\leq i\leq M)

0 \leq c_i \leq 1(1\leq i\leq M)

If i \neq j, then (a_i,b_i) \neq (a_j,b_j).

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1 c_1
:
a_M b_M c_M

Output

Print the number of possible ways to write integers, modulo 998244353.

Sample Input 1

3 3
1 1 1
3 1 0
2 3 1

Sample Output 1

8

For example, the following ways to write integers satisfy the condition:

101 111
011 111
000 011

Sample Input 2

4 5
1 3 1
2 4 0
2 3 1
4 2 1
4 4 1

Sample Output 2

32

Sample Input 3

3 5
1 3 1
3 3 0
3 1 0
2 3 1
3 2 1

Sample Output 3

0

Sample Input 4

4 8
1 1 1
1 2 0
3 2 1
1 4 0
2 1 1
1 3 0
3 4 1
4 4 1

Sample Output 4

4

Sample Input 5

100000 0

Sample Output 5

342016343

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

//author: camypaper
using System;
using System.Linq;
using System.Collections.Generic;
using System.Text.RegularExpressions;
using Debug = System.Diagnostics.Debug;
using SB = System.Text.StringBuilder;
using System.Numerics;
using Point = System.Numerics.Complex;
using static System.Math;
using C = System.Int32;

namespace Program {
    public class Solver {
        Random rnd = new Random(0);
        public void Solve() {
            var n = ri;
            var m = ri;
            var k = 1L * n * (n + 1) / 2 - n - (n - 1) - (n - 2);
            var s = Enumerate(2 * n, x => new List<int>());
            var t = Enumerate(n, x => new List<int>());

            var A = Enumerate(m, x => new int[] { ri - 1, ri - 1, ri });
            for (int i = 0; i < m; i++)
                if (A[i][0] > A[i][1]) Swap(ref A[i][0], ref A[i][1]);
            A = A.OrderBy(x => x[0]).ThenBy(x => x[1]).ToArray();
            var ok = true;
            for (int i = 0; i < m; i++) {
                var x = A[i][0]; var y = A[i][1];
                Debug.WriteLine(A[i].AsJoinedString());
                if (Abs(x - y) <= 1) s[x + y].Add(A[i][2]);
                else if (Abs(x - y) == 2) t[(x + y) / 2].Add(A[i][2]);
                else {
                    if (i + 1 < m && x == A[i + 1][0] && y == A[i + 1][1]) {
                        ok &= A[i][2] == A[i + 1][2];
                    }
                }
            }
            k -= A.Where(a => Abs(a[0] - a[1]) > 2).Select(a => a[0] * 1000000L + a[1]).Distinct().Count();
            Debug.WriteLine(k);
            if (!ok) { Console.WriteLine(0); return; }
            var dp = new ModInt[2];
            dp[0] = 1;
            for (int i = 0; i < n; i++) {
                Debug.WriteLine(dp.AsJoinedString());
                var next = new ModInt[2];
                for (int p = 0; p < 2; p++)
                    for (int q = 0; q < 2; q++) {
                        if (dp[p].num == 0) continue;
                        var coef = 1;
                        if (s[i + i].Count > 0 && s[i + i][0] != q) coef *= 0;
                        if (i > 0 && i + 1 < n) {
                            if (t[i].Count == 2) {
                                if (q == 0 && t[i].Sum() % 2 != 0) coef *= 0;
                                if (q == 1 && t[i].Sum() != 1) coef *= 0;
                            }
                            else if (t[i].Count == 0) coef *= 2;
                        }
                        if (i > 0) {
                            if (s[i + i - 1].Count == 2) {
                                if ((p ^ q ^ s[i + i - 1][0] ^ s[i + i - 1][1]) != 0) coef *= 0;
                            }
                            else if (s[i + i - 1].Count == 0) coef *= 2;

                        }
                        next[q] += coef * dp[p];
                    }

                dp = next;
            }
            Debug.WriteLine(dp.AsJoinedString());
            var ans = dp[0] + dp[1];
            ans *= ModInt.Pow(2, k);
            Console.WriteLine(ans);

        }


        const long INF = 1L << 60;
        int ri => sc.Integer();
        long rl => sc.Long();
        double rd => sc.Double();
        string rs => sc.Scan();
        public IO.StreamScanner sc = new IO.StreamScanner(Console.OpenStandardInput());

        static IEnumerable<int> Rep(int n) => Enumerable.Range(0, n);
        static IEnumerable<int> RRep(int n) => Enumerable.Range(0, n).Reverse();
        static T[] Enumerate<T>(int n, Func<int, T> f) {
            var a = new T[n];
            for (int i = 0; i < a.Length; ++i) a[i] = f(i);
            return a;
        }
        static public void Swap<T>(ref T a, ref T b) { var tmp = a; a = b; b = tmp; }
    }
}

#region main
static class Ex {
    static public string AsString(this IEnumerable<char> ie) { return new string(ie.ToArray()); }
    static public string AsJoinedString<T>(this IEnumerable<T> ie, string st = " ") {
        return string.Join(st, ie);
    }
    static public void Main() {
        Console.SetOut(new Program.IO.Printer(Console.OpenStandardOutput()) { AutoFlush = false });
        var solver = new Program.Solver();
        solver.Solve();
        Console.Out.Flush();
    }
}
#endregion
#region Ex
namespace Program.IO {
    using System.IO;
    using System.Text;
    using System.Globalization;

    public class Printer : StreamWriter {
        public override IFormatProvider FormatProvider => CultureInfo.InvariantCulture;
        public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
    }

    public class StreamScanner {
        public StreamScanner(Stream stream) { str = stream; }

        public readonly Stream str;
        private readonly byte[] buf = new byte[1024];
        private int len, ptr;
        public bool isEof = false;

        private byte read() {
            if (isEof) return 0;
            if (ptr >= len) {
                ptr = 0;
                if ((len = str.Read(buf, 0, 1024)) <= 0) {
                    isEof = true;
                    return 0;
                }
            }
            return buf[ptr++];
        }

        public char Char() {
            byte b = 0;
            do b = read(); while ((b < 33 || 126 < b) && !isEof);
            return (char)b;
        }
        public string Scan() {
            var sb = new StringBuilder();
            for (var b = Char(); b >= 33 && b <= 126; b = (char)read()) sb.Append(b);
            return sb.ToString();
        }
        public string ScanLine() {
            var sb = new StringBuilder();
            for (var b = Char(); b != '\n' && b != 0; b = (char)read()) if (b != '\r') sb.Append(b);
            return sb.ToString();
        }
        public long Long() { return isEof ? long.MinValue : long.Parse(Scan()); }
        public int Integer() { return isEof ? int.MinValue : int.Parse(Scan()); }
        public double Double() { return isEof ? double.NaN : double.Parse(Scan(), CultureInfo.InvariantCulture); }
    }
}

#endregion

#region ModInt
/// <summary>
/// [0,<see cref="Mod"/>) までの値を取るような数
/// </summary>
public struct ModInt {
    /// <summary>
    /// 剰余を取る値．
    /// </summary>
    public const long Mod = 998244353;

    /// <summary>
    /// 実際の数値．
    /// </summary>
    public long num;
    /// <summary>
    /// 値が <paramref name="n"/> であるようなインスタンスを構築します．
    /// </summary>
    /// <param name="n">インスタンスが持つ値</param>
    /// <remarks>パフォーマンスの問題上，コンストラクタ内では剰余を取りません．そのため，<paramref name="n"/> ∈ [0,<see cref="Mod"/>) を満たすような <paramref name="n"/> を渡してください．このコンストラクタは O(1) で実行されます．</remarks>
    public ModInt(long n) { num = n; }
    /// <summary>
    /// このインスタンスの数値を文字列に変換します．
    /// </summary>
    /// <returns>[0,<see cref="Mod"/>) の範囲内の整数を 10 進表記したもの．</returns>
    public override string ToString() { return num.ToString(); }
    public static ModInt operator +(ModInt l, ModInt r) { l.num += r.num; if (l.num >= Mod) l.num -= Mod; return l; }
    public static ModInt operator -(ModInt l, ModInt r) { l.num -= r.num; if (l.num < 0) l.num += Mod; return l; }
    public static ModInt operator *(ModInt l, ModInt r) { return new ModInt(l.num * r.num % Mod); }
    public static implicit operator ModInt(long n) { n %= Mod; if (n < 0) n += Mod; return new ModInt(n); }

    /// <summary>
    /// 与えられた 2 つの数値からべき剰余を計算します．
    /// </summary>
    /// <param name="v">べき乗の底</param>
    /// <param name="k">べき指数</param>
    /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
    public static ModInt Pow(ModInt v, long k) { return Pow(v.num, k); }

    /// <summary>
    /// 与えられた 2 つの数値からべき剰余を計算します．
    /// </summary>
    /// <param name="v">べき乗の底</param>
    /// <param name="k">べき指数</param>
    /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
    public static ModInt Pow(long v, long k) {
        long ret = 1;
        for (k %= Mod - 1; k > 0; k >>= 1, v = v * v % Mod)
            if ((k & 1) == 1) ret = ret * v % Mod;
        return new ModInt(ret);
    }
    /// <summary>
    /// 与えられた数の逆元を計算します．
    /// </summary>
    /// <param name="v">逆元を取る対象となる数</param>
    /// <returns>逆元となるような値</returns>
    /// <remarks>法が素数であることを仮定して，フェルマーの小定理に従って逆元を O(log N) で計算します．</remarks>
    public static ModInt Inverse(ModInt v) { return Pow(v, Mod - 2); }
}
#endregion