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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MaxN = 3e5 + 11;
int f[MaxN];
int dep[MaxN];
int fa[MaxN];
int id[MaxN];
int U[MaxN];
int V[MaxN];
int W[MaxN];
int N, M;
vector<pair<int, int> > adj[MaxN];
void dfs(int u, int pre) {
  int si = int(adj[u].size());
  fa[u] = pre, dep[u] = dep[pre] + 1;
  for (int i = 0; i < si; ++i) {
    int v = adj[u][i].first;
    if (v == pre) continue;
    dfs(v, u), id[v] = adj[u][i].second;
  }
}
void modify(int u, int v, int w) {
  if (dep[u] < dep[v]) swap(u, v);
  while (dep[u] > dep[v]) {
    f[id[u]] = max(f[id[u]], w);
    u = fa[u];
  }
  while (u != v) {
    f[id[u]] = max(f[id[u]], w), f[id[v]] = max(f[id[v]], w), u = fa[u],
    v = fa[v];
  }
}
bool query(int u, int v, int w) {
  int res = 2e9;
  if (dep[u] < dep[v]) swap(u, v);
  while (dep[u] > dep[v]) {
    res = min(f[id[u]], res);
    if (res < w) return false;
    u = fa[u];
  }
  while (u != v) {
    res = min(f[id[u]], res), res = min(f[id[v]], res);
    if (res < w) return false;
    u = fa[u], v = fa[v];
  }
  return w == res;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> N;
  for (int i = 1; i < N; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(make_pair(v, i)), adj[v].push_back(make_pair(u, i));
  }
  dfs(1, 0);
  cin >> M;
  for (int i = 1; i <= M; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    modify(u, v, w);
    U[i] = u, V[i] = v, W[i] = w;
  }
  bool ok = 1;
  for (int i = 1; i <= M; ++i) {
    ok &= query(U[i], V[i], W[i]);
    if (!ok) break;
  }
  if (!ok)
    puts("-1");
  else
    for (int i = 1; i < N; ++i) cout << (f[i] == 0 ? 1 : f[i]) << ' ';
  return 0;
}