Problem: There are n railway stations in Berland. They are connected to each other by n-1 railway sections. The railway network is connected, i.e. can be represented as an undirected tree.

You have a map of that network, so for each railway section you know which stations it connects.

Each of the n-1 sections has some integer value of the scenery beauty. However, these values are not marked on the map and you don't know them. All these values are from 1 to 10^6 inclusive.

You asked m passengers some questions: the j-th one told you three values:

  * his departure station a_j; 
  * his arrival station b_j; 
  * minimum scenery beauty along the path from a_j to b_j (the train is moving along the shortest path from a_j to b_j). 



You are planning to update the map and set some value f_i on each railway section — the scenery beauty. The passengers' answers should be consistent with these values.

Print any valid set of values f_1, f_2, ..., f_{n-1}, which the passengers' answer is consistent with or report that it doesn't exist.

Input

The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of railway stations in Berland.

The next n-1 lines contain descriptions of the railway sections: the i-th section description is two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), where x_i and y_i are the indices of the stations which are connected by the i-th railway section. All the railway sections are bidirected. Each station can be reached from any other station by the railway.

The next line contains a single integer m (1 ≤ m ≤ 5000) — the number of passengers which were asked questions. Then m lines follow, the j-th line contains three integers a_j, b_j and g_j (1 ≤ a_j, b_j ≤ n; a_j ≠ b_j; 1 ≤ g_j ≤ 10^6) — the departure station, the arrival station and the minimum scenery beauty along his path.

Output

If there is no answer then print a single integer -1.

Otherwise, print n-1 integers f_1, f_2, ..., f_{n-1} (1 ≤ f_i ≤ 10^6), where f_i is some valid scenery beauty along the i-th railway section.

If there are multiple answers, you can print any of them.

Examples

Input


4
1 2
3 2
3 4
2
1 2 5
1 3 3


Output


5 3 5


Input


6
1 2
1 6
3 1
1 5
4 1
4
6 1 3
3 4 1
6 5 2
1 2 5


Output


5 3 1 2 1 


Input


6
1 2
1 6
3 1
1 5
4 1
4
6 1 1
3 4 3
6 5 3
1 2 4


Output


-1

Solution (in PYTHON3):

Solution:
from collections import deque
from operator import itemgetter
import sys
input = sys.stdin.readline


def dfs(tree, root):
    par = {root: -1}
    st1 = deque([root])
    while st1:
        v = st1.pop()
        for nxt_v in tree[v]:
            if nxt_v in par:
                continue
            else:
                par[nxt_v] = v
                st1.append(nxt_v)
    return par

class LowestCommonAncestor():
    def __init__(self, tree, root):
        self.n = len(tree)
        self.depth = [0] * self.n
        self.log_size = (self.n).bit_length()
        self.parent = [[-1] * self.n for i in range(self.log_size)]

        q = deque([(root, -1, 0)]) 
        while q:
            v, par, dist = q.pop()
            self.parent[0][v] = par
            self.depth[v] = dist
            for child_v in tree[v]:
                if child_v != par:
                    self.depth[child_v] = dist + 1
                    q.append((child_v, v, dist + 1))

        for k in range(1, self.log_size):
            for v in range(self.n):
                self.parent[k][v] = self.parent[k-1][self.parent[k-1][v]]
            
    def lca(self, u, v):
        if self.depth[u] > self.depth[v]:
            u, v = v, u
        for k in range(self.log_size):
            if (self.depth[v] - self.depth[u] >> k) & 1:
                v = self.parent[k][v]
        if u == v:
            return u
          
        for k in reversed(range(self.log_size)):
            if self.parent[k][u] != self.parent[k][v]:
                u = self.parent[k][u]
                v = self.parent[k][v]
        return self.parent[0][u]


n = int(input())
info = [list(map(int, input().split())) for i in range(n - 1)]
m = int(input())
query = [list(map(int, input().split())) for i in range(m)]

tree = [[] for i in range(n)]
edge = {}
for i in range(n - 1):
    a, b = info[i]
    a -= 1
    b -= 1
    if a > b:
        a, b = b, a
    tree[a].append(b)
    tree[b].append(a)
    edge[a * 10000 + b] = i

root = 0
lca = LowestCommonAncestor(tree, root)
par = dfs(tree, root)

ans = [0] * (n - 1)
query = sorted(query, key=itemgetter(2))
for a, b, g in query:
    a -= 1
    b -= 1
    ab = lca.lca(a, b)
    set_ = set()
    while True:
        if a == ab:
            break
        ind = edge[min(a, par[a]) * 10000 + max(a, par[a])]
        ans[ind] = g
        a = par[a]
    while True:
        if b == ab:
            break
        ind = edge[min(b, par[b]) * 10000 + max(b, par[b])]
        ans[ind] = g
        b = par[b]
    for i in set_:
        ans[i] = g

for a, b, g in query:
    a -= 1
    b -= 1
    ab = lca.lca(a, b)
    res = 10 ** 10
    while True:
        if a == ab:
            break
        ind = edge[min(a, par[a]) * 10000 + max(a, par[a])]
        res = min(res, ans[ind])
        a = par[a]
    while True:
        if b == ab:
            break
        ind = edge[min(b, par[b]) * 10000 + max(b, par[b])]
        res = min(res, ans[ind])
        b = par[b]
    if res != g:
        print(-1)
        exit()

for i in range(n - 1):
    if ans[i] == 0:
        ans[i] = 10 ** 6
print(*ans)