'Twas the night before Christmas, and Santa's frantically setting up his new Christmas tree! There are n nodes in the tree, connected by n-1 edges. On each edge of the tree, there's a set of Christmas lights, which can be represented by an integer in binary representation.

<image>

He has m elves come over and admire his tree. Each elf is assigned two nodes, a and b, and that elf looks at all lights on the simple path between the two nodes. After this, the elf's favorite number becomes the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of the values of the lights on the edges in that path.

However, the North Pole has been recovering from a nasty bout of flu. Because of this, Santa forgot some of the configurations of lights he had put on the tree, and he has already left the North Pole! Fortunately, the elves came to the rescue, and each one told Santa what pair of nodes he was assigned (a_i, b_i), as well as the parity of the number of set bits in his favorite number. In other words, he remembers whether the number of 1's when his favorite number is written in binary is odd or even.

Help Santa determine if it's possible that the memories are consistent, and if it is, remember what his tree looked like, and maybe you'll go down in history!

Input

The first line contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t cases follow.

The first line of each test case contains two integers, n and m (2 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ m ≤ 2 ⋅ 10^5) — the size of tree and the number of elves respectively.

The next n-1 lines of each test case each contains three integers, x, y, and v (1 ≤ x, y ≤ n; -1 ≤ v < 2^{30}) — meaning that there's an edge between nodes x and y. If 

  * v = -1: Santa doesn't remember what the set of lights were on for this edge. 
  * v ≥ 0: The set of lights on the edge is v. 



The next m lines of each test case each contains three integers, a, b, and p (1 ≤ a, b ≤ n; a ≠ b; 0 ≤ p ≤ 1) — the nodes that the elf was assigned to, and the parity of the number of set bits in the elf's favorite number.

It is guaranteed that the sum of all n and the sum of all m don't exceed 2 ⋅ 10^5 each.

It is guaranteed that the given edges form a tree.

Output

For each test case, first print either YES or NO (in any case), whether there's a tree consistent with Santa's memory or not. 

If the answer is YES, print n-1 lines each containing three integers: x, y, and v (1 ≤ x, y ≤ n; 0 ≤ v < 2^{30}) — the edge and the integer on that edge. The set of edges must be the same as in the input, and if the value of some edge was specified earlier, it can not change. You can print the edges in any order.

If there are multiple answers, print any.

Example

Input


4
6 5
1 2 -1
1 3 1
4 2 7
6 3 0
2 5 -1
2 3 1
2 5 0
5 6 1
6 1 1
4 5 1
5 3
1 2 -1
1 3 -1
1 4 1
4 5 -1
2 4 0
3 4 1
2 3 1
3 3
1 2 -1
1 3 -1
1 2 0
1 3 1
2 3 0
2 1
1 2 1
1 2 0


Output


YES
1 2 0
1 3 1
2 4 7
3 6 0
2 5 0
YES
1 2 1
1 3 0
1 4 1
4 5 1
NO
NO

Note

The first test case is the image in the statement.

One possible answer is assigning the value of the edge (1, 2) to 5, and the value of the edge (2, 5) to 3. This is correct because: 

  * The first elf goes from node 2 to node 3. This elf's favorite number is 4, so he remembers the value 1 (as 4 has an odd number of 1 bits in its binary representation). 
  * The second elf goes from node 2 to node 5. This elf's favorite number is 3, so he remembers the value 0 (as 3 has an even number of 1 bits in its binary representation). 
  * The third elf goes from node 5 to node 6. This elf's favorite number is 7, so he remembers the value 1 (as 7 has an odd number of 1 bits in its binary representation). 
  * The fourth elf goes from node 6 to node 1. This elf's favorite number is 1, so he remembers the value 1 (as 1 has an odd number of 1 bits in its binary representation). 
  * The fifth elf goes from node 4 to node 5. This elf's favorite number is 4, so he remembers the number 1 (as 4 has an odd number of 1 bits in its binary representation). 



Note that there are other possible answers.
import java.io.*;
import java.util.*;

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

public class cf1615d {

    public static void main(String[] args) throws IOException {
        int t = ri();
        next: while (t --> 0) {
            int n = rni(), m = ni();
            dsu dsu = new dsu(2 * n);
            List<int[]> edges = new ArrayList<>();
            for (int i = 0; i < n - 1; ++i) {
                int x = rni() - 1, y = ni() - 1, v = ni();
                edges.add(new int[] {x, y, v});
                if (v != -1) {
                    int p = Integer.bitCount(v) & 1;
                    if (p == 0) {
                        dsu.union(x, y);
                        dsu.union(n + x, n + y);
                    } else {
                        dsu.union(x, n + y);
                        dsu.union(n + x, y);
                    }
                }
            }
            for (int i = 0; i < m; ++i) {
                int a = rni() - 1, b = ni() - 1, p = ni();
                if (p == 0) {
                    dsu.union(a, b);
                    dsu.union(n + a, n + b);
                } else {
                    dsu.union(a, n + b);
                    dsu.union(n + a, b);
                }
            }
            for (int i = 0; i < n; ++i) {
                if (dsu.find(i) == dsu.find(n + i)) {
                    prN();
                    continue next;
                }
            }
            prY();
            for (int[] e : edges) {
                if (e[2] == -1) {
                    prln(e[0], e[1], dsu.find(e[0]) == dsu.find(e[1]) ? 0 : 1);
                } else {
                    prln(e);
                }
            }
        }
        close();
    }

    static class dsu {
        boolean init = true;
        int n, par[], sz[];

        dsu(int n_) {
            n = n_;
            par = new int[n];
            sz = new int[n];
            if (init) {
                init();
            } else {
                fill(par, -1);
            }
        }

        void make(int v) {
            par[v] = v;
            sz[v] = 1;
        }

        void init() {
            for (int i = 0; i < n; ++i) {
                make(i);
            }
        }

        int find(int v) {
            if (par[v] == -1) {
                return -1;
            } else if (v == par[v]) {
                return v;
            } else {
                par[v] = find(par[v]);
                return par[v];
            }
        }

        void union(int u, int v) {
            int a = find(u), b = find(v);
            if (a != b) {
                if (sz[a] < sz[b]) {
                    int swap = a;
                    a = b;
                    b = swap;
                }
                par[b] = a;
                sz[a] += sz[b];
            }
        }
    }

    static BufferedReader __i = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter __o = new PrintWriter(new OutputStreamWriter(System.out));
    static StringTokenizer input;
    static Random __r = new Random();

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

    // constants
    static final int IBIG = 1000000007;
    static final int IMAX = 2147483647;
    static final long LMAX = 9223372036854775807L;
    // 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 gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}
    static long gcd(long a, long b) {return b == 0 ? a : gcd(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 randInt(int min, int max) {return __r.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 void ria(int[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = ni();}
    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 void riam1(int[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1;}
    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 void rla(long[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = nl();}
    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 void rda(double[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = nd();}
    static char[] rcha() throws IOException {return rline().toCharArray();}
    static void rcha(char[] a) throws IOException {int n = a.length, i = 0; for (char c : rline().toCharArray()) a[i++] = c;}
    static String rline() throws IOException {return __i.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) {__o.print(i);}
    static void prln(int i) {__o.println(i);}
    static void pr(long l) {__o.print(l);}
    static void prln(long l) {__o.println(l);}
    static void pr(double d) {__o.print(d);}
    static void prln(double d) {__o.println(d);}
    static void pr(char c) {__o.print(c);}
    static void prln(char c) {__o.println(c);}
    static void pr(char[] s) {__o.print(new String(s));}
    static void prln(char[] s) {__o.println(new String(s));}
    static void pr(String s) {__o.print(s);}
    static void prln(String s) {__o.println(s);}
    static void pr(Object o) {__o.print(o);}
    static void prln(Object o) {__o.println(o);}
    static void prln() {__o.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() {__o.flush();}
    static void close() {__o.close();}
}