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 power(long long int x, long long int y, long long int p) {
  long long int r = 1;
  x = x % p;
  while (y) {
    if (y & 1) r = r * x % p;
    y = y >> 1;
    x = x * x % p;
  }
  return r;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int csrand(int l = 0, int r = 1e9) {
  uniform_int_distribution<int> uid(l, r);
  return uid(rng);
}
const int N = 5e3 + 99;
vector<pair<int, int> > g[N];
int par[N], ht[N], ed[N], ans[N];
void dfs(int v, int p) {
  par[v] = p;
  for (auto x : g[v]) {
    if (x.first != p) {
      ht[x.first] = ht[v] + 1;
      ed[x.first] = x.second;
      dfs(x.first, v);
    }
  }
}
struct anjn {
  int u, v, wt;
} ARR[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) ans[i] = 1;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back({y, i});
    g[y].push_back({x, i});
  }
  dfs(1, 1);
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int u, v, wt;
    cin >> u >> v >> wt;
    ARR[i].u = u;
    ARR[i].v = v;
    ARR[i].wt = wt;
    while (ht[u] > ht[v]) {
      ans[ed[u]] = max(ans[ed[u]], wt);
      u = par[u];
    }
    while (ht[u] < ht[v]) {
      ans[ed[v]] = max(ans[ed[v]], wt);
      v = par[v];
    }
    while (u != v) {
      ans[ed[u]] = max(ans[ed[u]], wt);
      u = par[u];
      ans[ed[v]] = max(ans[ed[v]], wt);
      v = par[v];
    }
  }
  for (int i = 0; i < m; i++) {
    int u, v, wt;
    int ans1 = 1e7;
    u = ARR[i].u;
    v = ARR[i].v;
    wt = ARR[i].wt;
    while (ht[u] > ht[v]) {
      ans1 = min(ans[ed[u]], ans1);
      u = par[u];
    }
    while (ht[u] < ht[v]) {
      ans1 = min(ans[ed[v]], ans1);
      v = par[v];
    }
    while (u != v) {
      ans1 = min(ans[ed[u]], ans1);
      u = par[u];
      ans1 = min(ans[ed[v]], ans1);
      v = par[v];
    }
    if (ans1 != wt) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) cout << ans[i] << " ";
  cout << '\n';
}