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;
long long int em(long long int x, long long int Mod, long long int n) {
  if (n == 0)
    return 1ll;
  else if (n % 2 == 0)
    return em((x % Mod * x % Mod) % Mod, Mod, n / 2) % Mod;
  else
    return (x % Mod * em((x % Mod * x % Mod) % Mod, Mod, (n - 1) / 2)) % Mod;
}
vector<long long int> v[5009];
long long int beauty[5009], lim;
long long int tin[5009], tout[5009], tim;
vector<vector<long long int>> ancestor;
long long int edge[5009][5009];
void dfs(long long int x, long long int par) {
  tin[x] = tim++;
  ancestor[x][0] = par;
  edge[x][0] = par;
  for (long long int i = (long long int)1; i < (long long int)lim + 1; i++)
    ancestor[x][i] = ancestor[ancestor[x][i - 1]][i - 1];
  for (long long int y : v[x]) {
    if (y == par) continue;
    dfs(y, x);
  }
  tout[x] = tim++;
}
bool isancestor(long long int x, long long int y) {
  return ((tin[x] <= tin[y]) && (tout[x] >= tout[y]));
}
long long int lca(long long int x, long long int y) {
  if (isancestor(x, y)) return x;
  if (isancestor(y, x)) return y;
  for (long long int i = lim; i >= 0; i--) {
    if (isancestor(ancestor[x][i], y)) continue;
    x = ancestor[x][i];
  }
  return ancestor[x][0];
}
int32_t main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, i, j, m, k, l;
  cin >> n;
  for (long long int k = (long long int)0; k < (long long int)n - 1; k++) {
    cin >> i >> j;
    v[i].push_back(j);
    v[j].push_back(i);
    edge[i][j] = edge[j][i] = k;
  }
  vector<pair<long long int, pair<long long int, long long int>>> ques;
  cin >> m;
  while (m--) {
    cin >> i >> j >> l;
    ques.push_back({l, {i, j}});
  }
  sort(ques.begin(), ques.end());
  lim = ceil(log2(n));
  tim = 0;
  ancestor.assign(n + 1, vector<long long int>(lim + 1));
  dfs(1, 1);
  for (pair<long long int, pair<long long int, long long int>> y : ques) {
    long long int num = y.first;
    long long int x = y.second.first;
    long long int z = y.second.second;
    long long int lc = lca(x, z);
    while (x != lc) {
      long long int par = ancestor[x][0];
      beauty[edge[x][par]] = num;
      x = par;
    }
    while (z != lc) {
      long long int par = ancestor[z][0];
      beauty[edge[z][par]] = num;
      z = par;
    }
  }
  long long int maxi = (ques.back()).first;
  for (long long int k = (long long int)0; k < (long long int)n - 1; k++) {
    if (!beauty[k]) beauty[k] = maxi;
  }
  long long int flag = 0;
  for (pair<long long int, pair<long long int, long long int>> y : ques) {
    long long int mini = INT_MAX;
    long long int num = y.first;
    long long int x = y.second.first;
    long long int z = y.second.second;
    long long int lc = lca(x, z);
    while (x != lc) {
      long long int par = ancestor[x][0];
      mini = min(mini, beauty[edge[x][par]]);
      x = par;
    }
    while (z != lc) {
      long long int par = ancestor[z][0];
      mini = min(mini, beauty[edge[z][par]]);
      z = par;
    }
    if (num != mini) {
      flag = 1;
      break;
    }
  }
  if (flag)
    cout << "-1\n";
  else {
    for (long long int i = (long long int)0; i < (long long int)n - 1; i++)
      cout << beauty[i] << " ";
  }
}