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>
void err() { std::cout << '\n'; }
template <typename T, typename... Args>
void err(T a, Args... args) {
  std::cout << a << ' ';
  err(args...);
}
void pt() { std::cout << '\n'; }
template <typename T, typename... Args>
void pt(T a, Args... args) {
  std::cout << a << ' ';
  pt(args...);
}
using namespace std;
const int N = 5000 + 5;
int n, m;
vector<int> G[N];
int deep[N], fa[N];
int f[N], e[N][N], id[N][N];
int a[N], b[N], c[N];
void dfs(int u, int p) {
  deep[u] = deep[p] + 1;
  fa[u] = p;
  for (auto v : G[u]) {
    if (v != p) dfs(v, u);
  }
}
void dfs2(int u, int p) {
  for (auto v : G[u]) {
    if (v != p) {
      dfs2(v, u);
      f[id[u][v]] = e[u][v];
    }
  }
}
void run() {
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
    id[u][v] = id[v][u] = i;
  }
  dfs(1, 0);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d%d%d", &a[i], &b[i], &c[i]);
  for (int i = 1; i <= m; i++) {
    if (deep[a[i]] < deep[b[i]]) swap(a[i], b[i]);
    int k1 = a[i], k2 = b[i], g = c[i];
    while (deep[k1] != deep[k2]) {
      int to = fa[k1];
      if (e[to][k1] <= g) e[to][k1] = e[k1][to] = g;
      k1 = to;
    }
    while (k1 != k2) {
      int to1 = fa[k1], to2 = fa[k2];
      if (e[to1][k1] <= g) e[to1][k1] = e[k1][to1] = g;
      if (e[to2][k2] <= g) e[to2][k2] = e[k2][to2] = g;
      k1 = to1, k2 = to2;
    }
  }
  for (int i = 1; i <= m; i++) {
    if (deep[a[i]] < deep[b[i]]) swap(a[i], b[i]);
    int k1 = a[i], k2 = b[i], g = c[i];
    int Min = 0x3f3f3f3f;
    while (deep[k1] != deep[k2]) {
      int to = fa[k1];
      Min = min(Min, e[to][k1]);
      k1 = to;
    }
    while (k1 != k2) {
      int to1 = fa[k1], to2 = fa[k2];
      Min = min(Min, min(e[to1][k1], e[to2][k2]));
      k1 = to1, k2 = to2;
    }
    if (Min != g) {
      puts("-1");
      return;
    }
  }
  dfs2(1, 0);
  for (int i = 1; i < n; i++)
    if (!f[i]) f[i] = 1000000;
  for (int i = 1; i < n; i++) printf("%d ", f[i]);
}
int main() {
  while (~scanf("%d", &n)) run();
  return 0;
}