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 long long oo = 1e8;
const double pi = 3.1415926535897;
const double EPS = (1e-7);
int dcmp(double x, double y) { return fabs(x - y) <= EPS ? 0 : x < y ? -1 : 1; }
struct edge {
  int from, to, w;
  edge() {}
  edge(int x, int y, int we) : from(x), to(y), w(we) {}
  bool operator<(const edge &rhs) const { return w > rhs.w; }
};
const int N = 5005;
vector<int> adj[N];
int adjM[N][N];
int ancs[N], depth[N], par[N];
void dfs(int nd, int p, int hd, int d) {
  for (auto ch : adj[nd]) {
    if (ch == p) continue;
    int head = (nd == 1 ? ch : hd);
    ancs[ch] = head;
    depth[ch] = d;
    par[ch] = nd;
    dfs(ch, nd, head, d + 1);
  }
}
int up(int nd, int g, int lim, int ch) {
  int minA = 1e6 + 5;
  while (true) {
    int p = par[nd];
    if (nd == lim) break;
    int mini = min(p, nd);
    int maxi = max(p, nd);
    if (ch) {
      if (adjM[mini][maxi] == 0) adjM[mini][maxi] = g;
    }
    nd = p;
    minA = min(minA, adjM[mini][maxi]);
  }
  return minA;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> edgeL;
  int x, y;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
    int mini = min(x, y);
    int maxi = max(x, y);
    edgeL.push_back({mini, maxi});
  }
  dfs(1, 1, 1, 0);
  int m;
  cin >> m;
  int g;
  vector<pair<int, pair<int, int>>> vec;
  for (int i = 0; i < m; i++) {
    cin >> x >> y >> g;
    pair<int, pair<int, int>> q = make_pair(g, make_pair(x, y));
    vec.push_back(q);
  }
  sort(((vec).rbegin()), ((vec).rend()));
  int ch = 1;
  for (int it = 0; it < 2; it++) {
    if (it) {
    }
    for (int i = 0; i < m; i++) {
      auto q = vec[i];
      g = q.first;
      x = q.second.first;
      y = q.second.second;
      if (ancs[x] != ancs[y]) {
        int mini = up(x, g, 1, ch);
        int mini2 = up(y, g, 1, ch);
        mini = min(mini, mini2);
        if (ch == 0 && x != 1 && mini != g) return cout << -1, 0;
        if (ch == 0 && y != 1 && mini != g) return cout << -1, 0;
      } else {
        int minA = 1e6 + 5;
        if (depth[x] != depth[y]) {
          int nd = (depth[x] > depth[y] ? x : y);
          int lim = (depth[x] < depth[y] ? x : y);
          int need = depth[nd] - depth[lim];
          int tmp = nd;
          while (need--) {
            tmp = par[tmp];
          }
          minA = up(nd, g, tmp, ch);
          x = tmp;
          y = lim;
        }
        while (true) {
          int p1 = par[x];
          int p2 = par[y];
          if (x == y) break;
          int mini1 = min(p1, x);
          int maxi1 = max(p1, x);
          int mini2 = min(p2, y);
          int maxi2 = max(p2, y);
          if (ch) {
            if (adjM[mini1][maxi1] == 0) adjM[mini1][maxi1] = g;
            if (adjM[mini2][maxi2] == 0) adjM[mini2][maxi2] = g;
          }
          x = p1;
          y = p2;
          minA = min(minA, adjM[mini1][maxi1]);
          minA = min(minA, adjM[mini2][maxi2]);
        }
        if (ch == 0 && minA != g) return cout << -1, 0;
      }
    }
    ch ^= 1;
  }
  for (auto e : edgeL) {
    adjM[e.first][e.second] = max(1, adjM[e.first][e.second]);
    cout << adjM[e.first][e.second] << " ";
  }
}