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 = 5000 + 5;
int T, n, m;
vector<int> G[maxn];
int fa[maxn], dep[maxn];
int f[maxn];
int fr[maxn], to[maxn];
int ind[maxn][maxn];
bool solve(int x, int y, int v) {
  if (dep[x] > dep[y]) swap(x, y);
  int ok = 0;
  while (dep[x] != dep[y]) {
    if (!f[ind[y][fa[y]]] || f[ind[y][fa[y]]] == v) {
      f[ind[y][fa[y]]] = v;
      ok++;
    };
    y = fa[y];
  }
  if (x == y) return ok;
  while (x != y) {
    if (!f[ind[y][fa[y]]] || f[ind[y][fa[y]]] == v) {
      f[ind[y][fa[y]]] = v;
      ok++;
    };
    if (!f[ind[x][fa[x]]] || f[ind[x][fa[x]]] == v) {
      f[ind[x][fa[x]]] = v;
      ok++;
    };
    y = fa[y];
    x = fa[x];
  }
  return ok;
}
void dfs(int x) {
  for (auto v : G[x])
    if (v != fa[x]) {
      fa[v] = x;
      dep[v] = dep[x] + 1;
      dfs(v);
    }
}
struct Path {
  int x, y, z, i;
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    fr[i] = x;
    to[i] = y;
    ind[x][y] = ind[y][x] = i;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  dfs(1);
  cin >> m;
  int ok = 1;
  vector<Path> query;
  for (int i = 1; i <= m; ++i) {
    int x, y, z;
    cin >> x >> y >> z;
    query.push_back({x, y, z, i});
  }
  sort(query.begin(), query.end(), [](Path a, Path b) { return a.z > b.z; });
  for (auto [x, y, z, i] : query) {
    if (!solve(x, y, z)) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 1; i <= n - 1; ++i) {
    if (f[i])
      cout << f[i] << " ";
    else
      cout << 1 << " ";
  }
  cout << endl;
  return 0;
}