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 <bits/stdc++.h>

using namespace std;
 
using ll = long long;
using ld = long double;
using db = double; 
using str = string; // yay python!
 
using pi = pair<int,int>;
using pl = pair<ll,ll>;
using pd = pair<db,db>;
 
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>; 
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>; 
using vpd = vector<pd>;
 
#define tcT template<class T
#define tcTU tcT, class U
// ^ lol this makes everything look weird but I'll try it
tcT> using V = vector<T>; 
tcT, size_t SZ> using AR = array<T,SZ>; 
tcT> using PR = pair<T,T>;
 
// pairs
#define mp make_pair
#define f first
#define s second
 
// vectors
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define sor(x) sort(all(x)) 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back()
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 
 
// loops
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)
 
const int MOD = 1e9+7; // 998244353;
const int mxN = 2e2 + 5;
const int INF = 1e9; // not too close to LLONG_MAX
const ld PI = acos((ld)-1);
const int tSZ = (1 << 21);
const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; // for every grid problem!!
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 
template<class T> using pqg = priority_queue<T,vector<T>,greater<T>>;
 
// helper funcs
constexpr int pct(int x) { return __builtin_popcount(x); } // # of bits set
constexpr int bits(int x) { return 31-__builtin_clz(x); } // floor(log2(x)) 
ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down

ll add(ll f, ll s){ f += s; if(f >= MOD) f %= MOD; if(f < 0) f += MOD;	return f; } // add f to s and make sure it's in range
ll mul(ll f, ll s){ f *= s; if(f >= MOD) f %= MOD; return f;} // multiply f and s and make sure it's in range
ll pw(ll x, ll y) { ll ans = 1, ml = x; while(y) { if(y & 1) ans = mul(ans, ml); ml = mul(ml, ml);	y >>= 1; } return ans; } // number x to a power y
ll inv(ll x) { return pw(x, MOD - 2); }

tcT> bool ckmin(T& a, const T& b) {
	return b < a ? a = b, 1 : 0; } // set a = min(a,b)
tcT> bool ckmax(T& a, const T& b) {
	return a < b ? a = b, 1 : 0; }
 
tcTU> T fstTrue(T lo, T hi, U f) {
	hi ++; assert(lo <= hi); // assuming f is increasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo)/2;
		f(mid) ? hi = mid : lo = mid+1; 
	} 
	return lo;
}
tcTU> T lstTrue(T lo, T hi, U f) {
	lo --; assert(lo <= hi); // assuming f is decreasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo+1)/2;
		f(mid) ? lo = mid : hi = mid-1;
	} 
	return lo;
}
tcT> void remDup(vector<T>& v) { // sort and remove duplicates
	sort(all(v)); v.erase(unique(all(v)),end(v)); }
tcTU> void erase(T& t, const U& u) { // don't erase
	auto it = t.find(u); assert(it != end(t));
	t.erase(u); } // element that doesn't exist from (multi)set
 
 // INPUT
#define tcTUU tcT, class ...U
tcT> void re(complex<T>& c);
tcTU> void re(pair<T,U>& p);
tcT> void re(vector<T>& v);
tcT, size_t SZ> void re(AR<T,SZ>& a);
 
tcT> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
tcTUU> void re(T& t, U&... u) { re(t); re(u...); }
 
tcT> void re(complex<T>& c) { T a,b; re(a,b); c = {a,b}; }
tcTU> void re(pair<T,U>& p) { re(p.f,p.s); }
tcT> void re(vector<T>& x) { trav(a,x) re(a); }
tcT, size_t SZ> void re(AR<T,SZ>& x) { trav(a,x) re(a); }
tcT> void rv(int& n, vector<T>& x) { re(n); x.rsz(n); trav(a,x) re(a); }
 
// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(bool b) { 
	#ifdef LOCAL
		return b ? "true" : "false"; 
	#else 
		return ts((int)b);
	#endif
}
tcT> str ts(complex<T> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(vector<bool> v) {
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
tcTU> str ts(pair<T,U> p);
tcT> str ts(T v) { // containers with begin(), end()
	#ifdef LOCAL
		bool fst = 1; str res = "{";
		for (const auto& x: v) {
			if (!fst) res += ", ";
			fst = 0; res += ts(x);
		}
		res += "}"; return res;
	#else
		bool fst = 1; str res = "";
		for (const auto& x: v) {
			if (!fst) res += " ";
			fst = 0; res += ts(x);
		}
		return res;
 
	#endif
}
tcTU> str ts(pair<T,U> p) {
	#ifdef LOCAL
		return "("+ts(p.f)+", "+ts(p.s)+")"; 
	#else
		return ts(p.f)+" "+ts(p.s);
	#endif
}
 
// OUTPUT
tcT> void pr(T x) { cout << ts(x); }
tcTUU> void pr(const T& t, const U&... u) { 
	pr(t); pr(u...); }
void ps() { pr("\n"); } // print w/ spaces
tcTUU> void ps(const T& t, const U&... u) { 
	pr(t); if (sizeof...(u)) pr(" "); ps(u...); }
 
 // DEBUG
void DBG() { cerr << "]" << endl; }
tcTUU> void DBG(const T& t, const U&... u) {
	cerr << ts(t); if (sizeof...(u)) cerr << ", ";
	DBG(u...); }
#ifdef LOCAL // compile with -DLOCAL, chk -> fake assert
	#define dbg(...) cerr << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
	#define chk(...) if (!(__VA_ARGS__)) cerr << "Line(" << __LINE__ << ") -> function(" \
		 << __FUNCTION__  << ") -> CHK FAILED: (" << #__VA_ARGS__ << ")" << "\n", exit(0);
#else
	#define dbg(...) 0
	#define chk(...) 0
#endif
 
// FILE I/O
void setIn(str s) { freopen(s.c_str(),"r",stdin); }
void setOut(str s) { freopen(s.c_str(),"w",stdout); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
void setIO(str s = "") {
	unsyncIO();
	// cin.exceptions(cin.failbit); 
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}

int n, m, col[mxN];
vpi grph[mxN];

void nn()
{
	ps("NO");
	exit(0);
}

void check(int v, int p, int th)
{
	if(col[v] && col[v] != th)
	{
		//ps("NO");
		//exit();
		nn();
	}
	else if(col[v]) return;
	col[v] = th;
	trav(x, grph[v])
	{
		if(x.f != p)
		{
			check(x.f, v, 3 - th);
		}
	}
}
int dst[mxN];

bool btr()
{
	bool dn = 0;
	for(int i = 1; i <= n; i++)
	{
		trav(x, grph[i])
		{
			if(dst[x.f] > dst[i] + x.s)
			{
				dn = 1;
				dst[x.f] = dst[i] + x.s;
			}
		}
	}
	return dn;
}

int best, ans[mxN];

void trr(int v)
{
	for(int i = 1; i <= n; i++)
	{
		dst[i] = INF;
	}
	dst[v] = 0;
	for(int i = 1; i <= n + 3; i++)
	{
		btr();
	}
	if(btr())
	{
		nn();
	}
	int mn = 0;
	for(int i = 1; i <= n; i++)
	{
		mn = min(mn, dst[i]);
	}
	int bst = 0;
	for(int i = 1; i <= n; i++)
	{
		dst[i] += mn;
		bst = max(bst, dst[i]);
	}
	if(bst > best)
	{
		best = bst;
		for(int i = 1; i <= n; i++)
		{
			ans[i] = dst[i];
		}
	}
}

void solve()
{
	check(1, 1, 1);

	for(int i = 1; i <= n; i++)
	{
		trr(i);
	}

	ps("YES");
	ps(best);
	for(int i = 1; i <= n; i++)
	{
		pr(ans[i], " ");
	}
	ps();
}
 
int main()
{
	setIO();

	re(n, m);

	int v, w, c;
	FOR(i, 0, m)
	{
		re(v, w, c);
		if(c == 0)
		{
			grph[v].pb(mp(w, 1));
			grph[w].pb(mp(v, 1));
		}
		else
		{
			grph[v].pb(mp(w, 1));
			grph[w].pb(mp(v, -1));
		}
	}
	solve();
 
	return 0;
	// you should actually read the stuff at the bottom
}
 
 /* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
*/