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 = 5e3 + 10;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
int n, m;
vector<pair<int, int> > g[maxn];
struct edge {
  int u, v, w;
} e[maxn];
bool cmp(edge x, edge y) { return x.w < y.w; }
int pre[maxn][maxn][2];
int ans[maxn];
void dfs(int u, int fa, int root, int id) {
  pre[root][u][0] = id;
  pre[root][u][1] = fa;
  for (int i = 0; i < (int)g[u].size(); i++) {
    int v = g[u][i].first, t = g[u][i].second;
    if (v == fa) continue;
    dfs(v, u, root, t);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(pair<int, int>(v, i));
    g[v].push_back(pair<int, int>(u, i));
  }
  for (int i = 1; i <= n; i++) {
    dfs(i, 0, i, 0);
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
  }
  sort(e, e + m, cmp);
  for (int i = 0; i < m; i++) {
    int u = e[i].u, v = e[i].v, w = e[i].w;
    for (int j = v; j != u; j = pre[u][j][1]) {
      int id = pre[u][j][0];
      ans[id] = w;
    }
  }
  for (int i = 0; i < m; i++) {
    int u = e[i].u, v = e[i].v, w = e[i].w;
    int mi = inf;
    for (int j = v; j != u; j = pre[u][j][1]) {
      int id = pre[u][j][0];
      if (ans[id] == 0)
        ans[id] = 1e6;
      else {
        mi = min(mi, ans[id]);
      }
    }
    if (mi != w) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    if (ans[i] == 0) ans[i] = 1e6;
    printf("%d%c", ans[i], i == n - 1 ? '\n' : ' ');
  }
  return 0;
}