You are given a connected weighted undirected graph without any loops and multiple edges. 

Let us remind you that a graph's spanning tree is defined as an acyclic connected subgraph of the given graph that includes all of the graph's vertexes. The weight of a tree is defined as the sum of weights of the edges that the given tree contains. The minimum spanning tree (MST) of a graph is defined as the graph's spanning tree having the minimum possible weight. For any connected graph obviously exists the minimum spanning tree, but in the general case, a graph's minimum spanning tree is not unique.

Your task is to determine the following for each edge of the given graph: whether it is either included in any MST, or included at least in one MST, or not included in any MST.

Input

The first line contains two integers n and m (2 ≤ n ≤ 105, <image>) — the number of the graph's vertexes and edges, correspondingly. Then follow m lines, each of them contains three integers — the description of the graph's edges as "ai bi wi" (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106, ai ≠ bi), where ai and bi are the numbers of vertexes connected by the i-th edge, wi is the edge's weight. It is guaranteed that the graph is connected and doesn't contain loops or multiple edges.

Output

Print m lines — the answers for all edges. If the i-th edge is included in any MST, print "any"; if the i-th edge is included at least in one MST, print "at least one"; if the i-th edge isn't included in any MST, print "none". Print the answers for the edges in the order in which the edges are specified in the input.

Examples

Input

4 5
1 2 101
1 3 100
2 3 2
2 4 2
3 4 1


Output

none
any
at least one
at least one
any


Input

3 3
1 2 1
2 3 1
1 3 2


Output

any
any
none


Input

3 3
1 2 1
2 3 1
1 3 1


Output

at least one
at least one
at least one

Note

In the second sample the MST is unique for the given graph: it contains two first edges.

In the third sample any two edges form the MST for the given graph. That means that each edge is included at least in one MST.
from sys import stdin
from math import inf
from collections import defaultdict

    
class disjoinSet(object):
    def __init__(self,n):
        self.father = [x for x in range(0,n+1)]
        self.rank = [0 for x in range(0,n+1)]

    def setOf(self, x):
        if(self.father[x] != x):
            self.father[x] = self.setOf(self.father[x])
        return self.father[x]

    def Merge(self,x,y):
        xR = self.setOf(x)
        yR = self.setOf(y)
        if(xR == yR):
            return
        if self.rank[xR] < self.rank[yR]:
            self.father[xR] = yR
        elif self.rank[xR] > self.rank[yR]:
            self.father[yR] = xR
        else:
            self.father[yR] = xR
            self.rank[xR] +=1

##########

def mergesort(x):    
    if len(x) < 2:
        return x   
    middle = int(len(x)/2)    
    left = mergesort(x[:middle])
    right = mergesort(x[middle:])
    return merge(left,right)

def merge(left,right):
    result = []
    l,r=0,0
    while l < len(left) and r < len(right):        
        if left[l][2] <= right[r][2]:            
            result.append(left[l])
            l+=1            
        else:
            result.append(right[r])
            r+=1
    result += left[l:]
    result += right[r:]    
    return result

##########
    
def DFS(graph,bridges):
    time = 0
    for key in graph:
        low[key] = inf
        d[key] = -1
        pi[key] = key
        visited[key] = False
    for i in graph:
        if not visited[i]:
            DFS_Visit(graph,i,bridges)
    
def DFS_Visit(graph,u,bridges):
    visited[u] = True
    global time
    time +=1
    d[u] = time    
    low[u] = d[u]
    for v in graph[u]:
        if not visited[v]:
            pi[v] = u
            DFS_Visit(graph,v,bridges)
            low[u] = min(low[u],low[v])
        elif pi[u] != v:
            low[u] = min(low[u],d[v])
    if pi[u] != u and low[u] == d[u]:
        x = min(pi[u],u)
        y = max(pi[u],u)            
        bridges.append([x,y])
    time +=1

##########


n, m = map(int, stdin.readline().split())

if m > n-1:
    visited = {}
    d = {}
    low = {}
    pi = {}
    time = 0
    edges = []

    for i in range(0,m):
        x,y,z = map(int, stdin.readline().split())    
        edges.append([])
        edges[i] = [x,y,z]    

    ans = {}
    for i in edges:
        x = min(i[0],i[1])
        y = max(i[0],i[1])
        ans[x,y] = "at least one"

    dsu = disjoinSet(n)            

    sortEdges = mergesort(edges)

    criticalEdges = []
    i = 0
    cantEdges = m
    while i < cantEdges:        
        j = i;
        a = 0
        b = 0
        criticalEdges = []
        while j < len(sortEdges) and sortEdges[j][2] == sortEdges[i][2]:
            x = dsu.setOf(sortEdges[j][0])
            y = dsu.setOf(sortEdges[j][1])
            a = min(sortEdges[j][0],sortEdges[j][1])
            b = max(sortEdges[j][0],sortEdges[j][1])
            if x == y:            
                ans[a,b] = "none"
            criticalEdges.append(sortEdges[j])
            j+=1
        if len(criticalEdges) == 1:            
            if ans[a,b] != "none":
                ans[a,b] = "any"
                dsu.Merge(a,b)
        else:
            graph = defaultdict(set)
            crossCut = defaultdict(list)
            for m in criticalEdges:
                x = min(dsu.setOf(m[0]),dsu.setOf(m[1]))
                y = max(dsu.setOf(m[0]),dsu.setOf(m[1]))
                if x != y:                    
                    graph[x].add(y)
                    graph[y].add(x)
                    crossCut[x,y].append([min(m[0],m[1]),max(m[0],m[1])])
            bridges = []
            DFS(graph,bridges)
            for k in bridges:
                if len(crossCut[k[0],k[1]]) == 1:
                    x,y = crossCut[k[0],k[1]][0]                    
                    ans[x,y] = "any"
            
            for m in criticalEdges:
                dsu.Merge(m[0],m[1])                        
        i = j        
    for s in edges:        
        x = min(s[0],s[1])
        y = max(s[0],s[1])
        print(ans[x,y])
else:
     for _ in range(0,m):
        print("any")
