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\nsys.setrecursionlimit(10**6)\n\nclass UnionFind():\n    def __init__(self, n):\n        self.n = n\n        self.parents = [-1] * n\n\n    def find(self, x):\n        if self.parents[x] < 0:\n            return x\n        else:\n            self.parents[x] = self.find(self.parents[x])\n            return self.parents[x]\n\n    def union(self, x, y):\n        x = self.find(x)\n        y = self.find(y)\n\n        if x == y:\n            return\n\n        if self.parents[x] > self.parents[y]:\n            x, y = y, x\n\n        self.parents[x] += self.parents[y]\n        self.parents[y] = x\n\n    def size(self, x):\n        return -self.parents[self.find(x)]\n\n    def same(self, x, y):\n        return self.find(x) == self.find(y)\n\n    def members(self, x):\n        root = self.find(x)\n        return [i for i in range(self.n) if self.find(i) == root]\n\n    def roots(self):\n        return [i for i, x in enumerate(self.parents) if x < 0]\n\n    def group_count(self):\n        return len(self.roots())\n\n    def all_group_members(self):\n        return {r: self.members(r) for r in self.roots()}\n\n    def __str__(self):\n        return '\\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())\n\nn, m = map(int, input().split())\n\nans = [-1]*n\n\nuf = UnionFind(n)\ng = [[] for i in range(n)]\n\nfor i in range(m):\n    u, v, c = map(int, input().split())\n    if not uf.same(u-1, v-1):\n        uf.union(u-1, v-1)\n        g[u-1].append((v-1, c))\n        g[v-1].append((u-1, c))\n\ndef dfs(i):\n    for to, c in g[i]:\n        if ans[to]==-1:\n            if ans[i] == c:\n                if c == 1:\n                    ans[to] = c+1\n                else:\n                    ans[to] = c-1\n            else:\n                ans[to] = c\n            dfs(to)\n\nans = [-1]*n\nans[0] = 1\ndfs(0)\n\nif -1 in ans:\n    print('No')\n    return\n\nans = [a for a in ans]\n\nprint(*ans, sep='\\n')\n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: