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 int maxn = 3e5 + 5;
const int mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long ans = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b > 0 ? gcd(b, a % b) : a; }
int n, m, T;
int a[maxn];
vector<pair<int, int> > g[maxn];
int dep[maxn];
int f[maxn];
pair<int, int> val[maxn];
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  f[u] = fa;
  for (auto to : g[u]) {
    if (to.first == fa) continue;
    dfs(to.first, u);
    val[to.first] = {mod, to.second};
  }
}
struct node {
  int s, t, w;
  bool operator<(const node b) const { return w > b.w; }
} q[maxn];
bool gao(int x, int y, int w) {
  if (dep[x] < dep[y]) swap(x, y);
  int mi = mod;
  while (dep[x] > dep[y]) {
    if (val[x].first == mod) {
      val[x].first = w;
    }
    mi = min(val[x].first, mi);
    x = f[x];
  }
  while (x != y) {
    if (val[x].first == mod) {
      val[x].first = w;
    }
    if (val[y].first == mod) {
      val[y].first = w;
    }
    mi = min(val[x].first, mi);
    mi = min(val[y].first, mi);
    x = f[x];
    y = f[y];
  }
  return mi == w;
}
int ans[maxn];
int main() {
  cin >> n;
  for (int i = 1, uu, vv; i <= n - 1; i++) {
    cin >> uu >> vv;
    g[uu].push_back({vv, i});
    g[vv].push_back({uu, i});
  }
  dfs(1, 0);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> q[i].s >> q[i].t >> q[i].w;
  }
  sort(q + 1, q + 1 + m);
  for (int i = 1; i <= m; i++) {
    if (!gao(q[i].s, q[i].t, q[i].w)) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    ans[val[i].second] = val[i].first;
  }
  for (int i = 1; i <= n - 1; i++) {
    if (ans[i] == mod) ans[i] = 1e6;
    cout << ans[i] << " ";
  }
  return 0;
}