#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;
const long long MAXN = 5005;
vector<long long> edg[MAXN];
long long tin[MAXN];
long long tout[MAXN];
long long d = 0;
long long pred[MAXN];
long long wei[MAXN][MAXN];
long long INF = 1e6;
void dfs(long long v, long long p) {
  tin[v] = d++;
  pred[v] = p;
  for (auto u : edg[v]) {
    if (u == p) continue;
    dfs(u, v);
  }
  tout[v] = d++;
}
bool upper(long long u, long long v) {
  return tin[u] <= tin[v] && tout[u] >= tout[v];
}
bool cmp(tuple<long long, long long, long long> a,
         tuple<long long, long long, long long> b) {
  return get<2>(a) < get<2>(b);
}
void pol(long long a, long long b, long long w) {
  long long tmp = a;
  while (!upper(tmp, b)) {
    wei[tmp][pred[tmp]] = w;
    wei[pred[tmp]][tmp] = w;
    tmp = pred[tmp];
  }
  tmp = b;
  while (!upper(tmp, a)) {
    wei[tmp][pred[tmp]] = w;
    wei[pred[tmp]][tmp] = w;
    tmp = pred[tmp];
  }
}
long long check(long long a, long long b, long long w) {
  long long ans = INF, tmp = a;
  while (!upper(tmp, b)) {
    ans = min(ans, wei[tmp][pred[tmp]]);
    tmp = pred[tmp];
  }
  tmp = b;
  while (!upper(tmp, a)) {
    ans = min(ans, wei[tmp][pred[tmp]]);
    tmp = pred[tmp];
  }
  return ans == w;
}
void fast() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
signed main() {
  fast();
  long long n, i, a, b, m, w, j;
  cin >> n;
  vector<pair<long long, long long> > reb(n);
  for (i = 0; i < n - 1; i++) {
    cin >> a >> b;
    edg[a].push_back(b);
    edg[b].push_back(a);
    reb[i] = {a, b};
  }
  dfs(1, 0);
  cin >> m;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) wei[i][j] = INF;
  vector<tuple<long long, long long, long long> > zap(m);
  for (i = 0; i < m; i++) {
    cin >> a >> b >> w;
    zap[i] = make_tuple(a, b, w);
  }
  sort(zap.begin(), zap.end(), cmp);
  for (i = 0; i < m; i++) {
    tie(a, b, w) = zap[i];
    pol(a, b, w);
  }
  for (i = 0; i < m; i++) {
    tie(a, b, w) = zap[i];
    if (!check(a, b, w)) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (i = 0; i < n - 1; i++) {
    tie(a, b) = reb[i];
    cout << wei[a][b] << " ";
  }
}

### Prompt

In CPP, your task is to solve the following 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

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;
const long long MAXN = 5005;
vector<long long> edg[MAXN];
long long tin[MAXN];
long long tout[MAXN];
long long d = 0;
long long pred[MAXN];
long long wei[MAXN][MAXN];
long long INF = 1e6;
void dfs(long long v, long long p) {
  tin[v] = d++;
  pred[v] = p;
  for (auto u : edg[v]) {
    if (u == p) continue;
    dfs(u, v);
  }
  tout[v] = d++;
}
bool upper(long long u, long long v) {
  return tin[u] <= tin[v] && tout[u] >= tout[v];
}
bool cmp(tuple<long long, long long, long long> a,
         tuple<long long, long long, long long> b) {
  return get<2>(a) < get<2>(b);
}
void pol(long long a, long long b, long long w) {
  long long tmp = a;
  while (!upper(tmp, b)) {
    wei[tmp][pred[tmp]] = w;
    wei[pred[tmp]][tmp] = w;
    tmp = pred[tmp];
  }
  tmp = b;
  while (!upper(tmp, a)) {
    wei[tmp][pred[tmp]] = w;
    wei[pred[tmp]][tmp] = w;
    tmp = pred[tmp];
  }
}
long long check(long long a, long long b, long long w) {
  long long ans = INF, tmp = a;
  while (!upper(tmp, b)) {
    ans = min(ans, wei[tmp][pred[tmp]]);
    tmp = pred[tmp];
  }
  tmp = b;
  while (!upper(tmp, a)) {
    ans = min(ans, wei[tmp][pred[tmp]]);
    tmp = pred[tmp];
  }
  return ans == w;
}
void fast() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
signed main() {
  fast();
  long long n, i, a, b, m, w, j;
  cin >> n;
  vector<pair<long long, long long> > reb(n);
  for (i = 0; i < n - 1; i++) {
    cin >> a >> b;
    edg[a].push_back(b);
    edg[b].push_back(a);
    reb[i] = {a, b};
  }
  dfs(1, 0);
  cin >> m;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) wei[i][j] = INF;
  vector<tuple<long long, long long, long long> > zap(m);
  for (i = 0; i < m; i++) {
    cin >> a >> b >> w;
    zap[i] = make_tuple(a, b, w);
  }
  sort(zap.begin(), zap.end(), cmp);
  for (i = 0; i < m; i++) {
    tie(a, b, w) = zap[i];
    pol(a, b, w);
  }
  for (i = 0; i < m; i++) {
    tie(a, b, w) = zap[i];
    if (!check(a, b, w)) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (i = 0; i < n - 1; i++) {
    tie(a, b) = reb[i];
    cout << wei[a][b] << " ";
  }
}

```