Problem: A society can be represented by a connected, undirected graph of n vertices and m edges. The vertices represent people, and an edge (i,j) represents a friendship between people i and j.

In society, the i-th person has an income a_i. A person i is envious of person j if a_j=a_i+1. That is if person j has exactly 1 more unit of income than person i.

The society is called capitalist if for every pair of friends one is envious of the other. For some friendships, you know which friend is envious of the other. For the remaining friendships, you do not know the direction of envy.

The income inequality of society is defined as max_{1 ≤ i ≤ n} a_i - min_{1 ≤ i ≤ n} a_i.

You only know the friendships and not the incomes. If it is impossible for this society to be capitalist with the given knowledge, you should report about it. Otherwise, you should find an assignment of incomes in which the society is capitalist, and the income inequality is maximized.

Input

The first line contains two integers n, m (1≤ n≤ 200, n-1≤ m≤ 2000) — the number of people and friendships, respectively.

The following m lines describe the friendships. Each friendship is described by three integers i, j, b (1≤ i, j≤ n, i≠ j, 0≤ b≤ 1). This denotes that people i and j are friends. If b=1, we require that person i is envious of person j. If b=0, one friend should be envious of the other in either direction.

There is at most one friendship between each pair of people. It is guaranteed that if we consider the friendships as undirected edges, the graph is connected.

Output

Print "YES" if it is possible that the society is capitalist, or "NO" otherwise. You can print characters in any case (upper or lower).

If the answer is "YES", you should print two additional lines. In the first line, print the maximum possible income inequality. On the next line you should print n integers a_1,…, a_n (0≤ a_i≤ 10^6), where a_i denotes the income of the i-th person.

We can prove that if there exists a solution, there exists one where 0≤ a_i≤ 10^6 for all i.

If there exist multiple solutions, print any.

Examples

Input


6 6
1 2 0
3 2 0
2 5 0
6 5 1
6 3 0
2 4 1


Output


YES
3
3 2 1 3 1 0 


Input


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


Output


NO


Input


1 0


Output


YES
0
0 

Note

In the first test, we can show that an income inequality greater than 3 is impossible for the given society. In the given answer with income inequality equal to 3:

  * Person 2 is envious of person 1. 
  * Person 3 is envious of person 2. 
  * Person 5 is envious of person 2. 
  * Person 6 is envious of person 5 (the required direction is satisfied). 
  * Person 6 is envious of person 3. 
  * Person 2 is envious of person 4 (the required direction is satisfied). 



In the second test, we can show that there is no way to assign incomes to satisfy all requirements.

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>
#include<set>
#include<queue>
#include<map>
#include<algorithm>
#include<cstring>
#include<string>
#include<cassert>
#include<cmath>
#include<climits>
#include<iomanip>
#include<stack>
#include<unordered_map>
#include<bitset>
#include<limits>
#include<complex>
#include<array>
#include<numeric>
#include<functional>

using namespace std;
#define ll long long
#define ull unsigned long long
#define rrep(i,m,n) for(ll (i)=(ll)(m);(i)>=(ll)(n);(i)--)
#define rep(i,m,n) for(ll (i)=(ll)(m);i<(ll)(n);i++)
#define REP(i,n) rep(i,0,n)
#define FOR(i,c) for(decltype((c).begin())i=(c).begin();i!=(c).end();++i)
#define all(hoge) (hoge).begin(),(hoge).end()
typedef pair<ll, ll> P;
constexpr long double m_pi = 3.1415926535897932L;
constexpr ll MOD = 1000000007;
constexpr ll INF = 1LL << 61;
constexpr long double EPS = 1e-10;
template<typename T> using vector2 = vector<vector<T>>;
template<typename T> using vector3 = vector<vector2<T>>;
typedef vector<ll> Array;
typedef vector<Array> Matrix;
string operator*(const string& s, int k) {
	if (k == 0) return "";
	string p = (s + s) * (k / 2);
	if (k % 2 == 1) p += s;
	return p;
}

template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

struct Edge {//グラフ
	int to, rev; ll cap;
	Edge(int _to, ll _cap, int _rev) {
		to = _to; cap = _cap; rev = _rev;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;
void add_edge(Graph& G, int from, int to, ll cap, bool revFlag, ll revCap) {//最大フロー求める Ford-fulkerson
	G[from].push_back(Edge(to, cap, (ll)G[to].size() + (from == to)));
	if (revFlag)G[to].push_back(Edge(from, revCap, (ll)G[from].size() - 1));//最小カットの場合逆辺は0にする
}

ll max_flow_dfs(Graph& G, ll v, ll t, ll f, vector<bool>& used)
{
	if (v == t)
		return f;
	used[v] = true;
	for (int i = 0; i < G[v].size(); ++i) {
		Edge& e = G[v][i];
		if (!used[e.to] && e.cap > 0) {
			ll d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
			if (d > 0) {
				e.cap -= d;
				G[e.to][e.rev].cap += d;
				return d;
			}
		}
	}
	return 0;
}
//二分グラフの最大マッチングを求めたりも出来る　また二部グラフの最大独立集合は頂点数-最大マッチングのサイズ
ll max_flow(Graph& G, ll s, ll t)//O(V(V+E))
{
	ll flow = 0;
	for (;;) {
		vector<bool> used(G.size());
		REP(i, used.size())used[i] = false;
		ll f = max_flow_dfs(G, s, t, INF, used);
		if (f == 0) {
			return flow;
		}
		flow += f;
	}
}

void BellmanFord(Graph& G, ll s, Array& d, Array& negative) {//O(|E||V|)
	d.resize(G.size());
	negative.resize(G.size());
	REP(i, d.size())d[i] = INF;
	REP(i, d.size())negative[i] = false;
	d[s] = 0;
	REP(k, G.size() - 1) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
				}
			}
		}
	}
	REP(k, G.size() - 1) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
					negative[G[i][j].to] = true;
				}
				if (negative[i] == true)negative[G[i][j].to] = true;
			}
		}
	}
}
void Dijkstra(Graph& G, ll s, Array& d) {//O(|E|log|V|)
	d.resize(G.size());
	REP(i, d.size())d[i] = INF;
	d[s] = 0;
	priority_queue<P, vector<P>, greater<P>> q;
	q.push(make_pair(0, s));
	while (!q.empty()) {
		P a = q.top();
		q.pop();
		if (d[a.second] < a.first)continue;
		REP(i, G[a.second].size()) {
			Edge e = G[a.second][i];
			if (d[e.to] > d[a.second] + e.cap) {
				d[e.to] = d[a.second] + e.cap;
				q.push(make_pair(d[e.to], e.to));
			}
		}
	}
}
void WarshallFloyd(Graph& G, Matrix& d) {//O(V^3)
	d.resize(G.size());
	REP(i, d.size())d[i].resize(G.size());
	REP(i, d.size()) {
		REP(j, d[i].size()) {
			d[i][j] = ((i != j) ? INF : 0);
		}
	}
	REP(i, G.size()) {
		REP(j, G[i].size()) {
			chmin(d[i][G[i][j].to], G[i][j].cap);
		}
	}
	REP(i, G.size()) {
		REP(j, G.size()) {
			REP(k, G.size()) {
				chmin(d[j][k], d[j][i] + d[i][k]);
			}
		}
	}
}
bool tsort(Graph& graph, Array& order) {//トポロジカルソートO(E+V)
	int n = graph.size(), k = 0;
	Array in(n);
	for (auto& es : graph)
		for (auto& e : es)in[e.to]++;
	priority_queue<ll, Array, greater<ll>> que;
	REP(i, n)
		if (in[i] == 0)que.push(i);
	while (que.size()) {
		int v = que.top();
		que.pop();
		order.push_back(v);
		for (auto& e : graph[v])
			if (--in[e.to] == 0)que.push(e.to);
	}
	if (order.size() != n)return false;
	else return true;
}
class Lca {
public:
	const int n = 0;
	const int log2_n = 0;
	std::vector<std::vector<int>> parent;
	std::vector<int> depth;

	Lca() {}

	Lca(const Graph& g, int root)
		: n(g.size()), log2_n(log2(n) + 1), parent(log2_n, std::vector<int>(n)), depth(n) {
		dfs(g, root, -1, 0);
		for (int k = 0; k + 1 < log2_n; k++) {
			for (int v = 0; v < (int)g.size(); v++) {
				if (parent[k][v] < 0)
					parent[k + 1][v] = -1;
				else
					parent[k + 1][v] = parent[k][parent[k][v]];
			}
		}
	}

	void dfs(const Graph& g, int v, int p, int d) {
		parent[0][v] = p;
		depth[v] = d;
		for (auto& e : g[v]) {
			if (e.to != p) dfs(g, e.to, v, d + 1);
		}
	}

	int get(int u, int v) {
		if (depth[u] > depth[v]) std::swap(u, v);
		for (int k = 0; k < log2_n; k++) {
			if ((depth[v] - depth[u]) >> k & 1) {
				v = parent[k][v];
			}
		}
		if (u == v) return u;
		for (int k = log2_n - 1; k >= 0; k--) {
			if (parent[k][u] != parent[k][v]) {
				u = parent[k][u];
				v = parent[k][v];
			}
		}
		return parent[0][u];
	}
};

class UnionFind {
	vector<int> data;
	ll _num;
public:
	UnionFind(int size) : data(size, -1), _num(size) { }
	bool merge(int x, int y) {//xとyの集合を統合する
		x = root(x); y = root(y);
		if (x != y) {
			if (data[y] < data[x]) swap(x, y);
			data[x] += data[y]; data[y] = x;
		}
		_num -= (x != y);
		return x != y;
	}
	bool same(int x, int y) {//xとyが同じ集合か返す
		return root(x) == root(y);
	}
	int root(int x) {//xのルートを返す
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	ll size(int x) {//xの集合のサイズを返す
		return -data[root(x)];
	}
	ll num() {//集合の数を返す
		return _num;
	}
};

template<typename T, typename F>
class SegmentTree {
private:
	T identity;
	F merge;
	ll n;
	vector<T> dat;
public:
	SegmentTree(F f, T id, vector<T> v) :merge(f), identity(id) {
		int _n = v.size();
		n = 1;
		while (n < _n)n *= 2;
		dat.resize(2 * n - 1, identity);
		REP(i, _n)dat[n + i - 1] = v[i];
		for (int i = n - 2; i >= 0; i--)dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]);
	}
	SegmentTree(F f, T id, int _n) :merge(f), identity(id) {
		n = 1;
		while (n < _n)n *= 2;
		dat.resize(2 * n - 1, identity);
	}
	void set_val(int i, T x) {
		i += n - 1;
		dat[i] = x;
		while (i > 0) {
			i = (i - 1) / 2;
			dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]);
		}
	}
	T query(int l, int r) {
		T left = identity, right = identity;
		l += n - 1; r += n - 1;
		while (l < r) {
			if ((l & 1) == 0)left = merge(left, dat[l]);
			if ((r & 1) == 0)right = merge(dat[r - 1], right);
			l = l / 2;
			r = (r - 1) / 2;
		}
		return merge(left, right);
	}
};

template< typename T >
class FenwickTree {
	vector< T > data;
	int n;
	int p;
public:
	FenwickTree(int n) :n(n) {
		data.resize(n + 1LL, 0);
		p = 1;
		while (p < data.size())p *= 2;
	}
	T sum(int k) {
		T ret = 0;
		for (; k > 0; k -= k & -k) ret += data[k];
		return (ret);
	}

	T sum(int a, int b) { return sum(b) - sum(a); }//[a,b)

	void add(int k, T x) {
		for (++k; k <= n; k += k & -k) data[k] += x;
	}

	int lower_bound(ll w) {
		if (w <= 0)return -1;
		int x = 0;
		for (int k = p / 2; k > 0; k /= 2) {
			if (x + k <= n && data[x + k] < w)w -= data[x + k], x += k;
		}
		return x;
	}
};



//約数求める //約数
void divisor(ll n, vector<ll>& ret) {
	for (ll i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(ret.begin(), ret.end());
}

void prime_factorization(ll n, vector<P>& ret) {
	for (ll i = 2; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back({ i,0 });
			while (n % i == 0) {
				n /= i;
				ret[ret.size() - 1].second++;
			}
		}
	}
	if (n != 1)ret.push_back({ n,1 });
}


inline ll mod_pow(ll x, ll n, ll mod) {
	ll res = 1;
	while (n > 0) {
		if (n & 1) res = res * x % mod;
		x = x * x % mod;
		n >>= 1;
	}
	return res;
}

inline ll mod_inv(ll x, ll mod) {
	return mod_pow(x, mod - 2, mod);
}

class Combination {
public:
	Array fact;
	Array fact_inv;
	ll mod;
	//if n >= mod use lucas 
	ll nCr(ll n, ll r) {
		if (n < r)return 0;
		if (n < mod)return ((fact[n] * fact_inv[r] % mod) * fact_inv[n - r]) % mod;

		ll ret = 1;
		while (n || r) {
			ll _n = n % mod, _r = r % mod;
			n /= mod; r /= mod;
			(ret *= nCr(_n, _r)) %= mod;
		}
		return ret;
	}
	ll nPr(ll n, ll r) {
		return (fact[n] * fact_inv[n - r]) % mod;
	}
	ll nHr(ll n, ll r) {
		return nCr(r + n - 1, r);
	}
	Combination(ll _n, ll _mod) {
		mod = _mod;
		ll n = min(_n + 1, mod);
		fact.resize(n);
		fact[0] = 1;
		REP(i, n - 1) {
			fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
		}
		fact_inv.resize(n);
		fact_inv[n - 1] = mod_inv(fact[n - 1], mod);
		for (int i = n - 1; i > 0; i--) {
			fact_inv[i - 1] = fact_inv[i] * i % mod;
		}
	}
};

ll popcount(ll x) {
	x = (x & 0x5555555555555555) + (x >> 1 & 0x5555555555555555);
	x = (x & 0x3333333333333333) + (x >> 2 & 0x3333333333333333);
	x = (x & 0x0F0F0F0F0F0F0F0F) + (x >> 4 & 0x0F0F0F0F0F0F0F0F);
	x = (x & 0x00FF00FF00FF00FF) + (x >> 8 & 0x00FF00FF00FF00FF);
	x = (x & 0x0000FFFF0000FFFF) + (x >> 16 & 0x0000FFFF0000FFFF);
	x = (x & 0x00000000FFFFFFFF) + (x >> 32 & 0x00000000FFFFFFFF);

	return x;
}


template<typename T, typename E, typename F, typename G, typename H, typename P>
class LazySegmentTree {
private:
	const T identity_t;
	const E identity_e;
	F f_tt;
	G f_te;
	H f_ee;
	P f_ep;//区間に比例する場合
	vector<T> dat;
	vector<E> lazy;
	int n;
	void eval(int k, int len) {
		if (lazy[k] == identity_e)return;
		if (k < n - 1) {
			lazy[2 * k + 1] = f_ee(lazy[2 * k + 1], lazy[k]);
			lazy[2 * k + 2] = f_ee(lazy[2 * k + 2], lazy[k]);
		}
		dat[k] = f_te(dat[k], f_ep(lazy[k], len));
		lazy[k] = identity_e;
	}
	void update(int a, int b, E x, int k, int l, int r) {
		eval(k, r - l);
		if (a <= l && r <= b) {
			lazy[k] = f_ee(lazy[k], x);
			eval(k, r - l);
		}
		else if (a < r && l < b) {
			update(a, b, x, k * 2 + 1, l, (l + r) / 2);
			update(a, b, x, k * 2 + 2, (l + r) / 2, r);
			dat[k] = f_tt(dat[2 * k + 1], dat[2 * k + 2]);
		}
	}
	T query(int a, int b, int k, int l, int r) {
		eval(k, r - l);
		if (r <= a || b <= l)return identity_t;
		else if (a <= l && r <= b)return dat[k];
		else {
			T vl = query(a, b, 2 * k + 1, l, (l + r) / 2);
			T vr = query(a, b, 2 * k + 2, (l + r) / 2, r);
			return f_tt(vl, vr);
		}
	}
public:
	LazySegmentTree(vector<T> v, T id_t, E id_e, F f, G g, H h, P p) :
		identity_t(id_t), identity_e(id_e), f_tt(f), f_te(g), f_ee(h), f_ep(p) {
		n = 1;
		while (n < v.size())n *= 2;
		dat.resize(2 * n - 1, identity_t);
		lazy.resize(2 * n - 1, identity_e);
		REP(i, v.size())dat[n + i - 1] = v[i];
		for (int i = n - 2; i >= 0; i--)dat[i] = f_tt(dat[i * 2 + 1], dat[i * 2 + 2]);

	}
	void update(int a, int b, E x) { update(a, b, x, 0, 0, n); }
	T query(int a, int b) { return query(a, b, 0, 0, n); }
};



Matrix mIdentity(ll n) {
	Matrix A(n, Array(n));
	for (int i = 0; i < n; ++i) A[i][i] = 1;
	return A;
}

Matrix mMul(const Matrix& A, const Matrix& B,ll M) {
	Matrix C(A.size(), Array(B[0].size()));
	for (int i = 0; i < C.size(); ++i)
		for (int k = 0; k < A[i].size(); ++k)
			for (int j = 0; j < C[i].size(); ++j)
				(C[i][j] += (A[i][k] % M) * (B[k][j] % M)) %= M;
	return C;
}
// O( n^3 log e )
Matrix mPow(const Matrix& A, ll e , ll M) {
	return e == 0 ? mIdentity(A.size()) :
		e % 2 == 0 ? mPow(mMul(A, A, M), e / 2, M) : mMul(A, mPow(A, e - 1, M), M);
}


template<ll mod, ll root>//特殊な素数と原始根　998244353のとき3
class NTT {
private:
	template<typename T>
	inline void bit_reverse(vector<T>& a) {
		int n = a.size();
		int i = 0;
		for (int j = 1; j < n - 1; ++j) {
			for (int k = n >> 1; k > (i ^= k); k >>= 1);
			if (j < i) swap(a[i], a[j]);
		}
	}
	void _ntt(vector<long long>& a, int sign) {
		const int n = a.size();
		assert((n ^ (n & -n)) == 0); //n = 2^k

		long long tmp = (mod - 1) * mod_pow((ll)n, mod - 2, mod) % mod; // -1/n
		long long h = mod_pow(root, tmp, mod); // ^n√g
		if (sign == -1) h = mod_pow(h, mod - 2, mod);

		bit_reverse(a);

		for (ll m = 1; m < n; m <<= 1) {
			const ll m2 = 2 * m;
			long long _base = mod_pow((ll)h, (ll)(n / m2), mod);
			long long _w = 1;
			for (int x = 0; x < m; ++x) {
				for (ll s = x; s < n; s += m2) {
					long long u = a[s];
					long long d = (a[s + m] * _w) % mod;
					a[s] = (u + d) % mod;
					a[s + m] = (u - d + mod) % mod;
				}
				_w = (_w * _base) % mod;
			}
		}
	}

	void ntt(vector<long long>& input) { _ntt(input, 1); }

	void intt(vector<long long>& input) {
		_ntt(input, -1);
		const long long n_inv = mod_pow((ll)input.size(), mod - 2, mod);
		for (auto& x : input) x = (x * n_inv) % mod;
	}
public:
	// 畳み込み演算を行う
	vector<long long> convolution(const vector<long long>& a, const vector<long long>& b) {
		int result_size = a.size() + b.size() - 1;
		int n = 1; while (n < result_size) n <<= 1;

		vector<long long> _a = a, _b = b;
		_a.resize(n, 0);
		_b.resize(n, 0);

		ntt(_a);
		ntt(_b);
		for (int i = 0; i < n; ++i) _a[i] = (_a[i] * _b[i]) % mod;

		intt(_a);

		_a.resize(result_size);
		return _a;
	}
};


class StronglyConnectedComponents {
private:
	int n;
	Graph& g;
	void visit(int v, stack<int>& S, vector<int>& inS, vector<int>& low, vector<int>& num, int& time) {
		low[v] = num[v] = ++time;
		S.push(v); inS[v] = true;
		for (auto e : g[v]) {
			int w = e.to;
			if (num[w] == 0) {
				visit(w, S, inS, low, num, time);
				low[v] = min(low[v], low[w]);
			}
			else if (inS[w])
				low[v] = min(low[v], num[w]);
		}
		if (low[v] == num[v]) {
			scc.push_back(vector<int>());
			while (1) {
				int w = S.top(); S.pop(); inS[w] = false;
				scc.back().push_back(w);
				comp[w] = scc.size() - 1;
				if (v == w) break;
			}
		}
	}
public:
	vector2<int> scc;
	vector<int> comp;
	StronglyConnectedComponents(Graph& _g) :g(_g) {
		n = g.size();
		comp.resize(n);
		vector<int> num(n), low(n);
		stack<int> S;
		vector<int> inS(n);
		int time = 0;
		REP(u, n)if (num[u] == 0)visit(u, S, inS, low, num, time);
	}
	void build(Graph& t) {
		t.resize(scc.size());
		map<pair<int, int>, int> mp;
		REP(i, n) {
			int u = comp[i];
			for (auto e : g[i]) {
				int v = comp[e.to];
				if (u != v && mp.count({ u,v }) == 0)add_edge(t, u, v, 1, false, 0), mp[{u, v}] = 1;
			}
		}
	}
};



class HeavyLightDecomposition {
private:
	Graph& g;
	int n, root;
	vector<int> sz, par, in, out, head, rev, dep;
	void dfs_sz(int v, int p) {
		sz[v] = 1, par[v] = p;
		for (auto& e : g[v]) {
			if (e.to == p)swap(e, g[v].back());
			if (e.to == p)continue;
			dep[e.to] = dep[v] + 1;
			dfs_sz(e.to, v);
			sz[v] += sz[e.to];
			if (sz[e.to] > sz[g[v][0].to])swap(e, g[v][0]);
		}
	}
	void dfs_hld(int v, int p, int& t) {
		in[v] = t++; rev[in[v]] = v;
		for (auto e : g[v]) {
			if (e.to == p)continue;
			head[e.to] = (e.to == g[v][0].to ? head[v] : e.to);
			dfs_hld(e.to, v, t);
		}
		out[v] = t;
	}
public:
	HeavyLightDecomposition(Graph& _g, int _root = 0)
		:g(_g), n(g.size()), root(_root), sz(n), par(n, -1), head(n, -1), in(n), out(n), rev(n), dep(n, 0) {
		dfs_sz(root, -1);
		int t = 0;
		head[root] = root;
		dfs_hld(root, -1, t);
	}
	inline int get_in(int a) { return in[a]; }
	inline int get_sz(int a) { return sz[a]; }
	inline int get_rev(int a) { return rev[a]; }
	inline int get_dep(int a) { return dep[a]; }
	int lca(int a, int b) {
		for (;; b = par[head[b]]) {
			if (in[a] > in[b])swap(a, b);
			if (head[a] == head[b])return a;
		}
	}
	P subtree_query(int a) { return { in[a], out[a] }; }
	vector<P> path_query(int a, int b, bool edge = false) {
		vector<P> ret;
		for (;; b = par[head[b]]) {
			if (in[a] > in[b])swap(a, b);
			if (head[a] == head[b])break;
			ret.push_back({ in[head[b]],in[b] + 1 });
		}
		ret.push_back({ in[a] + edge,in[b] + 1 });
		return ret;
	}
};

template<typename E, typename H >
class DuelSegmentTree {
private:
	int sz, height;
	vector<E> lazy;
	const E identity_e;
	H f_ee;

	inline void propagate(int k) {
		if (lazy[k] != identity_e) {
			lazy[2 * k + 0] = f_ee(lazy[2 * k + 0], lazy[k]);
			lazy[2 * k + 1] = f_ee(lazy[2 * k + 1], lazy[k]);
			lazy[k] = identity_e;
		}
	}

	inline void thrust(int k) {
		for (int i = height; i > 0; i--) propagate(k >> i);
	}

public:
	DuelSegmentTree(int n, E id_e, H h) :
		identity_e(id_e), f_ee(h) {
		sz = 1;
		height = 0;
		while (sz < n) sz <<= 1, height++;
		lazy.resize(2 * sz, identity_e);
	}

	void update(int a, int b, const E& x) {
		thrust(a += sz);
		thrust(b += sz - 1);
		for (int l = a, r = b + 1; l < r; l >>= 1, r >>= 1) {
			if (l & 1) lazy[l] = f_ee(lazy[l], x), ++l;
			if (r & 1) --r, lazy[r] = f_ee(lazy[r], x);
		}
	}

	E operator[](int k) {
		thrust(k += sz);
		return lazy[k];
	}
};


template<typename T>
void rowReduction(vector<T> mat, vector<T>& basis) {//掃き出し法
	for (auto e : mat) {
		for (auto b : basis)
			chmin(e, e ^ b);
		if (e)
			basis.push_back(e);
	}
	sort(all(basis), greater<T>());
}

constexpr ll mod = 998244353;



int main() {
	ios::sync_with_stdio(false);
	std::cin.tie(0);
	std::cout.tie(0);


	ll n, m;
	cin >> n >> m;
	Graph g(n);
	UnionFind uni(2 * n);
	REP(i, m) {
		ll u, v, c;
		cin >> u >> v >> c;
		u--; v--;
		uni.merge(u, n + v);
		uni.merge(u + n, v);
		if (c == 0)add_edge(g, u, v, 1, true, 1);
		else add_edge(g, u, v, 1, true, -1);
	}
	REP(i, n) {
		if (uni.same(i, i + n)) {
			cout << "NO" << "\n";
			return 0;
		}
	}
	Matrix d;
	WarshallFloyd(g, d);
	REP(i, n) {
		if (d[i][i] < 0) {
			cout << "NO" << "\n";
			return 0;
		}
	}
	ll ans = -1;
	ll s;
	REP(i, n) {
		if (chmax(ans, *max_element(all(d[i]))))s = i;
	}
	cout << "YES" << "\n" << ans << "\n";
	REP(i, n)cout << d[s][i] << (i == n - 1 ? "\n" : " ");




	return 0;
}