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\n\nsys.setrecursionlimit(10 ** 7)\ninput = sys.stdin.readline\nf_inf = float(\'inf\')\nmod = 10 ** 9 + 7\n\n\nclass UnionFind:\n    def __init__(self, n):\n        self.n = n\n        self.parents = [-1] * n\n\n    def find(self, x):\n        """xの親を返す"""\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        """yをxの根に繋ぐ（マージテク有）"""\n        x = self.find(x)\n        y = self.find(y)\n        if x == y:\n            return\n        if self.parents[x] > self.parents[y]:\n            x, y = y, x\n        self.parents[x] += self.parents[y]\n        self.parents[y] = x\n\n    def same(self, x, y):\n        """xとyが同じ連結成分か判別する"""\n        return self.find(x) == self.find(y)\n\n    def size(self, x):\n        """xの連結成分の大きさを返す"""\n        return -self.parents[self.find(x)]\n\n    def kruskal(self, edge):\n        """\n        :param edge: edge = [(コスト, 頂点1, 頂点2),...]の形で重み付き隣接リストを渡して下さい\n        :return: 最小全域木のコストの和\n        """\n        edge.sort()\n        cost_sum = 0\n        for cost, node1, node2 in edge:\n            if not self.same(node1, node2):\n                cost_sum += cost\n                self.union(node1, node2)\n        return cost_sum\n\n\ndef resolve():\n    def dfs(v):\n        for d, u in edge[v]:\n            if uf.same(v, u):\n                continue\n            uf.union(u, v)\n            if res[v] != d:\n                res[u] = d\n            else:\n                for i in range(1, n + 1):\n                    if i != d:\n                        res[u] = i\n                        break\n            dfs(u)\n\n    n, m = map(int, input().split())\n    edge = [[] for _ in range(n)]\n    for _ in range(m):\n        u, v, c = map(int, input().split())\n        edge[u - 1].append([c, v - 1])\n        edge[v - 1].append([c, u - 1])\n\n    uf = UnionFind(n)\n    res = [0] * n\n    res[0] = 1\n    dfs(0)\n    print(*res, sep="\\n")\n\n\ndef __starting_point():\n    resolve()\n\n__starting_point()'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 18415