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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

random_device(rd);
mt19937 rng(rd());
const long long FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();

struct custom_hash
{
	template<class T>
	unsigned long long operator()(T v) const
	{
		unsigned long long x = v;
		x += FIXED_RANDOM;
		// x += 11400714819323198485ull;
		// x = (x ^ (x >> 30)) * 13787848793156543929ull;
		x = (x ^ (x >> 27)) * 10723151780598845931ull;
		return x ^ (x >> 31);
	}
};

template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T, class U> using hash_table = gp_hash_table<T, U, custom_hash>;

template<class T>
T randomize(T mod)
{
	return (uniform_int_distribution<T>(0, mod - 1))(rng);
}
template<class T>
void readi(T &x)
{
	x = 0;
	bool negative = false;
	char c = ' ';
	while (c < '-')
	{
		c = getchar();
	}
	if (c == '-')
	{
		negative = true;
		c = getchar();
	}
	while (c >= '0')
	{
		x = x * 10 + (c - '0');
		c = getchar();
	}
	if (negative)
	{
		x = -x;
	}
}
template<class T>
void printi(T output)
{
	if (output == 0)
	{
		putchar('0');
		return;
	}
	if (output < 0)
	{
		putchar('-');
		output = -output;
	}
	int buf[20], n = 0;
	while(output)
	{
		buf[n] = ((output % 10));
		output /= 10;
		n++;
	}
	for (n--; n >= 0; n--)
	{
		putchar(buf[n] + '0');
	}
	return;
}
template<class T>
void ckmin(T &a, T b)
{
	a = min(a, b);
}
template<class T>
void ckmax(T &a, T b)
{
	a = max(a, b);
}
long long expo(long long a, long long e, long long mod)
{
	return ((e == 0) ? 1 : ((expo(a * a % mod, e >> 1, mod)) * ((e & 1) ? a : 1) % mod));
}
template<class T, class U>
void nmod(T &x, U mod)
{
	if (x >= mod) x -= mod;
}
template<class T>
T gcd(T a, T b)
{
	return (b ? gcd(b, a % b) : a);
}

#define y0 ___y0
#define y1 ___y1
#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define DBG(x) cerr << #x << " = " << (x) << endl
#define SZ(x) ((int) ((x).size()))
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define ALL(x) (x).begin(), (x).end()

const long double PI = 4.0 * atan(1.0);
const long double EPS = 1e-9;

#define MAGIC 347
#define SINF 10007
#define CO 1000007
#define INF 998244353
#define BIG 1000000931
#define LARGE 1696969696967ll
#define GIANT 2564008813937411ll
#define LLINF 2696969696969696969ll
#define MAXN 100013

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ld, ld> pdd;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<ld> vd;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
typedef vector<pdd> vpd;

int N, M;
hash_table<int, bool> go[MAXN];
ll tot;
ll ans;

int fetch(int x, int y)
{
	if (go[x].find(y) == go[x].end()) return -1;
	return go[x][y];
}

int32_t main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	// cout << fixed << setprecision(10);
	// cerr << fixed << setprecision(10);
	// if (fopen("file.in", "r"))
	// {
	// 	freopen ("file.in", "r", stdin);
	// 	freopen ("file.out", "w", stdout);
	// }
	cin >> N >> M;
	tot = 1ll * N * (N + 1) / 2;
	FOR(i, 0, M)
	{
		int x, y, v;
		cin >> x >> y >> v;
		x--; y--;
		if (go[x].find(y) != go[x].end())
		{
			if (go[x][y] != v)
			{
				cout << "0\n";
				return 0;
			}
			continue;
		}
		tot--;
		go[x][y] = v;
		if (abs(x - y) >= 3)
		{
			go[y][x] = v;
			continue;
		}
		if (abs(x - y) == 2)
		{
			//u don't lose a degree of freedom!
			//(x, y), (y, x), (x + 1, y - 1)
			int v1 = fetch(y, x), v2 = fetch(min(x, y) + 1, max(x, y) - 1);
			if (v1 != -1)
			{
				go[min(x, y) + 1][max(x, y) - 1] = v ^ v1;
			}
			if (v2 != -1)
			{
				go[y][x] = v ^ v2;
			}
		}
		if (abs(x - y) == 1)
		{
			int v1 = fetch(y, x), v2 = fetch(x, x), v3 = fetch(y, y);
			if (v1 != -1 && v2 != -1)
			{
				go[y][y] = v ^ v1 ^ v2;
			}
			if (v1 != -1 && v3 != -1)
			{
				go[x][x] = v ^ v1 ^ v3;
			}
			if (v2 != -1 && v3 != -1)
			{
				go[y][x] = v ^ v2 ^ v3;
			}
		}
		if (abs(x - y) == 0)
		{
			if (x)
			{
				int v1 = fetch(x - 1, x - 1), v2 = fetch(x - 1, x), v3 = fetch(x, x - 1);
				if (v1 != -1 && v2 != -1)
				{
					go[x][x - 1] = v ^ v1 ^ v2;
				}
				if (v1 != -1 && v3 != -1)
				{
					go[x - 1][x] = v ^ v1 ^ v3;
				}
				if (v2 != -1 && v3 != -1)
				{
					go[x - 1][x - 1] = v ^ v2 ^ v3;
				}
			}
			if (x != N - 1)
			{
				int v1 = fetch(x, x + 1), v2 = fetch(x + 1, x), v3 = fetch(x + 1, x + 1);
				if (v1 != -1 && v2 != -1)
				{
					go[x + 1][x + 1] = v ^ v1 ^ v2;
				}
				if (v1 != -1 && v3 != -1)
				{
					go[x + 1][x] = v ^ v1 ^ v3;
				}
				if (v2 != -1 && v3 != -1)
				{
					go[x][x + 1] = v ^ v2 ^ v3;
				}
			}
		}
	}
	//now we find everything that's forced
	//if there are any contradictions, answer is 0, otherwise it's 2^(smth)
	tot %= (INF - 1);
	ans = expo(2, tot, INF);
	cout << ans << '\n';
	// cerr << "time elapsed = " << (clock() / (CLOCKS_PER_SEC / 1000)) << " ms" << endl;
	return 0;
}
/* READ READ READ
* int overflow, maxn too small, special cases (n=1?, two distinct?), cin.tie() interactive
* reread the problem, try small cases
* note down possible sources of error as you go
* do smth instead of nothing
*/