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 N = 5005;
const long long MOD = 1e9 + 7;
const long long INF = 1e9;
long long dp[N][32];
long long a[N];
long long h[N];
vector<long long> g[N];
long long tin[N], tout[N], timer = 0;
void dfs(long long v, long long p) {
  h[v] = h[p] + 1;
  tin[v] = timer++;
  dp[v][0] = p;
  for (long long i = 1; i < 32; ++i) dp[v][i] = dp[dp[v][i - 1]][i - 1];
  for (auto &it : g[v]) {
    if (it == p) continue;
    dfs(it, v);
  }
  tout[v] = timer++;
}
bool cmp(long long x, long long y) {
  return tin[x] <= tin[y] && tout[y] <= tout[x];
}
long long lca(long long x, long long y) {
  if (cmp(x, y)) return x;
  if (cmp(y, x)) return y;
  for (long long i = 30; i >= 0; --i) {
    if (!cmp(dp[x][i], y)) x = dp[x][i];
  }
  return dp[x][0];
}
bool check(long long x, long long y, long long val) {
  if (h[x] < h[y]) return false;
  while (x != y) {
    if (a[x] == 0 || a[x] == val) return true;
    x = dp[x][0];
    if (x == 0 || x == 1) break;
  }
  return false;
}
void upd(long long x, long long y, long long val) {
  if (h[x] < h[y]) return;
  while (x != y) {
    if (a[x] == 0) a[x] = val;
    x = dp[x][0];
    if (x == 0 || x == 1) break;
  }
}
mt19937_64 rng(time(0));
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<pair<int, int> > rd;
  for (long long i = 1; i < n; ++i) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
    rd.push_back({u, v});
  }
  dfs(1, 1);
  vector<pair<int, pair<int, int> > > ar;
  long long m;
  cin >> m;
  for (long long i = 0; i < m; ++i) {
    long long x, y, z;
    cin >> x >> y >> z;
    ar.push_back({z, {x, y}});
  }
  sort((ar).rbegin(), (ar).rend());
  for (auto &it : ar) {
    long long x = it.second.first, y = it.second.second;
    long long val = it.first;
    long long lc = lca(x, y);
    if (!check(x, lc, val) && !check(y, lc, val)) {
      cout << -1;
      return 0;
    }
    upd(x, lc, val);
    upd(y, lc, val);
  }
  for (auto &it : rd) {
    long long x = it.first, y = it.second;
    if (h[x] < h[y]) swap(x, y);
    if (a[x] != 0)
      cout << a[x] << " ";
    else
      cout << "1000000 ";
  }
  return 0;
}