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;
int i, i0, n, m, ans[5005];
vector<int> mp[5005];
struct pair_hash {
  inline size_t operator()(const pair<int, int>& p) const {
    return p.first * 1007 + p.second;
  }
};
unordered_map<pair<int, int>, int, pair_hash> dic;
struct node {
  int x, y, z;
} a[5005];
bool cmp(node a, node b) { return a.z < b.z; }
bool dfs(int x, int root, int y, int z) {
  if (y == x) return 1;
  for (int nex : mp[x]) {
    if (nex == root) continue;
    if (dfs(nex, x, y, z)) {
      ans[dic[{min(nex, x), max(nex, x)}]] = z;
      return 1;
    }
  }
  return 0;
}
int dfs0(int x, int root, int y, int z) {
  if (y == x) return INT_MAX - 1;
  for (int nex : mp[x]) {
    if (nex == root) continue;
    int vv = dfs0(nex, x, y, z);
    if (vv != INT_MAX) {
      return min(vv, ans[dic[{min(nex, x), max(nex, x)}]]);
    }
  }
  return INT_MAX;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (x > y) swap(x, y);
    dic[{x, y}] = i - 1;
    mp[x].push_back(y);
    mp[y].push_back(x);
    ans[i - 1] = 1;
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].z);
  }
  sort(a + 1, a + 1 + m, cmp);
  for (int i = 1; i <= m; i++) {
    dfs(a[i].x, -1, a[i].y, a[i].z);
  }
  for (int i = 1; i <= m; i++) {
    if (dfs0(a[i].x, -1, a[i].y, a[i].z) != a[i].z) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 1; i < n; i++) printf("%d ", ans[i]);
  return 0;
}