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 JAVA):

Solution:
import java.io.*;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

// editorial: make directed edges w weight 1 forward, -1 backward, undirected edges w weight 1 both ways then f-w

public class cf1450e {

    public static void main(String[] args) throws IOException {
        int n = rni(), m = ni(), dist[][] = new int[n][n], e[][] = new int[m][2];
        for (int[] row : dist) {
            fill(row, IBIG);
        }
        for (int i = 0; i < m; ++i) {
            int u = rni() - 1, v = ni() - 1, b = ni();
            e[i][0] = u;
            e[i][1] = v;
            dist[u][v] = 1;
            dist[v][u] = b == 0 ? 1 : -1;
        }
        for (int i = 0; i < n; ++i) {
            dist[i][i] = 0;
        }
        for (int k = 0; k < n; ++k) {
            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (dist[i][k] + dist[k][j] < dist[i][j]) {
                        dist[i][j] = dist[i][k] + dist[k][j];
                    }
                }
            }
        }
        // max, src
        int ans[] = {-1, -1};
        for (int i = 0; i < n; ++i) {
            if (dist[i][i] < 0) {
                // negative cycle
                prN();
                close();
                return;
            }
            for (int j = 0; j < n; ++j) {
                if (dist[i][j] > ans[0]) {
                    ans[0] = dist[i][j];
                    ans[1] = i;
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            if (dist[ans[1]][e[i][0]] == dist[ans[1]][e[i][1]]) {
                // odd cycle, not bipartite
                prN();
                close();
                return;
            }
        }
        prY();
        prln(ans[0]);
        prln(dist[ans[1]]);
        close();
    }

    static BufferedReader __in = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter __out = new PrintWriter(new OutputStreamWriter(System.out));
    static StringTokenizer input;
    static Random __rand = new Random();

    // references
    // IBIG = 1e9 + 7
    // IMAX ~= 2e9
    // LMAX ~= 9e18

    // constants
    static final int IBIG = 1000000007;
    static final int IMAX = 2147483647;
    static final int IMIN = -2147483648;
    static final long LMAX = 9223372036854775807L;
    static final long LMIN = -9223372036854775808L;
    // math util
    static int minof(int a, int b, int c) {return min(a, min(b, c));}
    static int minof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); int min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
    static long minof(long a, long b, long c) {return min(a, min(b, c));}
    static long minof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); long min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
    static int maxof(int a, int b, int c) {return max(a, max(b, c));}
    static int maxof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); int max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
    static long maxof(long a, long b, long c) {return max(a, max(b, c));}
    static long maxof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); long max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
    static int powi(int a, int b) {if (a == 0) return 0; int ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static long powl(long a, int b) {if (a == 0) return 0; long ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static int fli(double d) {return (int) d;}
    static int cei(double d) {return (int) ceil(d);}
    static long fll(double d) {return (long) d;}
    static long cel(double d) {return (long) ceil(d);}
    static int gcf(int a, int b) {return b == 0 ? a : gcf(b, a % b);}
    static long gcf(long a, long b) {return b == 0 ? a : gcf(b, a % b);}
    static int[] exgcd(int a, int b) {if (b == 0) return new int[] {1, 0}; int[] y = exgcd(b, a % b); return new int[] {y[1], y[0] - y[1] * (a / b)};}
    static long[] exgcd(long a, long b) {if (b == 0) return new long[] {1, 0}; long[] y = exgcd(b, a % b); return new long[] {y[1], y[0] - y[1] * (a / b)};}
    static int lcm(int a, int b) {return a * b / gcf(a, b);}
    static long lcm(long a, long b) {return a * b / gcf(a, b);}
    static int randInt(int min, int max) {return __rand.nextInt(max - min + 1) + min;}
    static long mix(long x) {x += 0x9e3779b97f4a7c15L; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L; x = (x ^ (x >> 27)) * 0x94d049bb133111ebL; return x ^ (x >> 31);}
    // array util
    static void reverse(int[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {int swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(long[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {long swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(double[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {double swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(char[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {char swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void shuffle(int[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void shuffle(long[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void shuffle(double[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void rsort(int[] a) {shuffle(a); sort(a);}
    static void rsort(long[] a) {shuffle(a); sort(a);}
    static void rsort(double[] a) {shuffle(a); sort(a);}
    static int[] copy(int[] a) {int[] ans = new int[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static long[] copy(long[] a) {long[] ans = new long[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static double[] copy(double[] a) {double[] ans = new double[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static char[] copy(char[] a) {char[] ans = new char[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    // input
    static void r() throws IOException {input = new StringTokenizer(rline());}
    static int ri() throws IOException {return Integer.parseInt(rline());}
    static long rl() throws IOException {return Long.parseLong(rline());}
    static double rd() throws IOException {return Double.parseDouble(rline());}
    static int[] ria(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni(); return a;}
    static int[] riam1(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1; return a;}
    static long[] rla(int n) throws IOException {long[] a = new long[n]; r(); for (int i = 0; i < n; ++i) a[i] = nl(); return a;}
    static double[] rda(int n) throws IOException {double[] a = new double[n]; r(); for (int i = 0; i < n; ++i) a[i] = nd(); return a;}
    static char[] rcha() throws IOException {return rline().toCharArray();}
    static String rline() throws IOException {return __in.readLine();}
    static String n() {return input.nextToken();}
    static int rni() throws IOException {r(); return ni();}
    static int ni() {return Integer.parseInt(n());}
    static long rnl() throws IOException {r(); return nl();}
    static long nl() {return Long.parseLong(n());}
    static double rnd() throws IOException {r(); return nd();}
    static double nd() {return Double.parseDouble(n());}
    // output
    static void pr(int i) {__out.print(i);}
    static void prln(int i) {__out.println(i);}
    static void pr(long l) {__out.print(l);}
    static void prln(long l) {__out.println(l);}
    static void pr(double d) {__out.print(d);}
    static void prln(double d) {__out.println(d);}
    static void pr(char c) {__out.print(c);}
    static void prln(char c) {__out.println(c);}
    static void pr(char[] s) {__out.print(new String(s));}
    static void prln(char[] s) {__out.println(new String(s));}
    static void pr(String s) {__out.print(s);}
    static void prln(String s) {__out.println(s);}
    static void pr(Object o) {__out.print(o);}
    static void prln(Object o) {__out.println(o);}
    static void prln() {__out.println();}
    static void pryes() {prln("yes");}
    static void pry() {prln("Yes");}
    static void prY() {prln("YES");}
    static void prno() {prln("no");}
    static void prn() {prln("No");}
    static void prN() {prln("NO");}
    static boolean pryesno(boolean b) {prln(b ? "yes" : "no"); return b;};
    static boolean pryn(boolean b) {prln(b ? "Yes" : "No"); return b;}
    static boolean prYN(boolean b) {prln(b ? "YES" : "NO"); return b;}
    static void prln(int... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static void prln(long... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static void prln(double... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for (int i = 0; i < n; pr(iter.next()), pr(' '), ++i); if (n >= 0) prln(iter.next()); else prln();}
    static void h() {prln("hlfd"); flush();}
    static void flush() {__out.flush();}
    static void close() {__out.close();}
}