Given is an undirected connected graph with N vertices numbered 1 to N, and M edges numbered 1 to M.
The given graph may contain multi-edges but not self loops.
Each edge has an integer label between 1 and N (inclusive).
Edge i has a label c_i, and it connects Vertex u_i and v_i bidirectionally.
Snuke will write an integer between 1 and N (inclusive) on each vertex (multiple vertices may have the same integer written on them) and then keep only the edges satisfying the condition below, removing the other edges.
Condition: Let x and y be the integers written on the vertices that are the endpoints of the edge. Exactly one of x and y equals the label of the edge.
We call a way of writing integers on the vertices good if (and only if) the graph is still connected after removing the edges not satisfying the condition above. Determine whether a good way of writing integers exists, and present one such way if it exists.

-----Constraints-----
 - 2 \leq N \leq 10^5
 - N-1 \leq M \leq 2 \times 10^5
 - 1 \leq u_i,v_i,c_i \leq N
 - The given graph is connected and has no self-loops.

-----Input-----
Input is given from Standard Input in the following format:
N M
u_1 v_1 c_1
\vdots
u_M v_M c_M

-----Output-----
If there is no good way of writing integers, print No.
Otherwise, print N lines. The i-th line should contain the integer written on Vertex i.
Any good way of writing integers will be accepted.

-----Sample Input-----
3 4
1 2 1
2 3 2
3 1 3
1 3 1

-----Sample Output-----
1
2
1

 - We write 1, 2, and 1 on Vertex 1, 2, and 3, respectively.
 - Edge 1 connects Vertex 1 and 2, and its label is 1.
 - Only the integer written on Vertex 1 equals the label, so this edge will not get removed.
 - Edge 2 connects Vertex 2 and 3, and its label is 2.
 - Only the integer written on Vertex 2 equals the label, so this edge will not be removed.
 - Edge 3 connects Vertex 1 and 3, and its label is 3.
 - Both integers written on the vertices differ from the label, so this edge will be removed.
 - Edge 4 connects Vertex 1 and 3, and its label is 1.
 - Both integers written on the vertices equal the label, so this edge will be removed.
 - After Edge 3 and 4 are removed, the graph will still be connected, so this is a good way of writing integers.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import sys\nfrom collections import deque\nfrom heapq import heappush, heappop\ndef input(): return sys.stdin.readline().strip()\n\ndef main():\n    N, M = list(map(int, input().split()))\n    repn = [[] for _ in range(N + 1)]\n    for _ in range(M):\n        u, v, c = list(map(int, input().split()))\n        repn[u].append((c, v))\n        repn[v].append((c, u))\n    """\n    グラフにおける構築問題、spanning treeを考えるとうまくいくのは王道なのか？\n    最小全域木はプリムでいいのかな、随分久しぶりに使う。\n\n    木が作れたら、根から順番にラベル付を行う。\n    例えばある親ノードpから辺ラベルcのついた辺で降りてきた頂点vを考えるとき、\n    頂点vにはラベルcをつける。そして子に順々と下っていく。\n    最初の根につけるラベルは、根から伸びるどの辺のラベルとも一致しないラベルを付ければ良い。\n    これは木の辺が頂点数より１小さいことから必ずできることが保証される！\n\n    注意点として、親の親から連続で辺のラベルが連続で一致する場合、この作戦だと\n    親と子で頂点のラベルが一致するので辺が切れてしまう。\n    なのでこの場合だけ子には親と違うラベルをランダムに当てる。\n    """\n    # Prim\n    visited = [0] * (N + 1)\n    used_edge = [[] for _ in range(N + 1)]\n    edge_q = []\n    for c, v in repn[1]:\n        heappush(edge_q, (c, 1, v))\n    visited[1] = 1\n\n    while edge_q:\n        c, u, v = heappop(edge_q)\n        if visited[v]: continue\n        visited[v] = 1\n        used_edge[u].append((c, v))\n        for nc, nv in repn[v]:\n            if not visited[nv]: heappush(edge_q, (nc, v, nv))\n        \n    # Labeling\n    label = [0] * (N + 1)\n    cand = set(range(1, N + 1))\n    for c, v in used_edge[1]: cand.discard(c)\n    label[1] = cand.pop()\n\n    q = deque()\n    for c, v in used_edge[1]: q.append((c, 1, v))\n    while q:\n        c, u, v = q.popleft()\n        if label[u] == c:\n            label[v] = 1 if c != 1 else 2\n        else:\n            label[v] = c\n        for nc, nv in used_edge[v]:\n            q.append((nc, v, nv))\n\n    for i, l in enumerate(label):\n        if i == 0: continue\n        print(l)\n\ndef __starting_point():\n    main()\n\n\n\n\n\n__starting_point()'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 16035