#include <bits/stdc++.h>
using namespace std;
const int sz = 5e3 + 9;
int n, m;
vector<pair<int, int>> v[sz];
int a, b, g;
int ans[sz];
int first, t;
set<pair<int, pair<int, int>>> mys;
int dfs1(int idx, int to, int val, int par = -1) {
  if (idx == to) return 1;
  int ret = 0;
  for (auto u : v[idx]) {
    int nxt = u.first;
    int i = u.second;
    if (nxt == par) continue;
    int tmp = dfs1(nxt, to, val, idx);
    if (tmp) ans[i] = val;
    ret |= tmp;
  }
  return ret;
}
int dfs2(int idx, int to, int par = -1) {
  if (idx == to) return -1;
  int ret = 1e9;
  for (auto u : v[idx]) {
    int nxt = u.first;
    int i = u.second;
    if (nxt == par) continue;
    int tmp = dfs2(nxt, to, idx);
    if (tmp != 1e9) {
      if (~tmp)
        ret = min(ret, min(tmp, ans[i]));
      else
        ret = min(ret, ans[i]);
    }
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d %d", &first, &t);
    v[first].push_back(make_pair(t, i));
    v[t].push_back(make_pair(first, i));
    ans[i] = 1e9;
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &a, &b, &g);
    mys.insert(make_pair(g, make_pair(a, b)));
  }
  for (auto u : mys) {
    g = u.first;
    a = u.second.first;
    b = u.second.second;
    dfs1(a, b, g);
  }
  for (auto u : mys) {
    g = u.first;
    a = u.second.first;
    b = u.second.second;
    int tmp = dfs2(a, b);
    if (tmp != g) {
      puts("-1");
      return 0;
    }
  }
  int mx = 1e6;
  for (int i = 1; i <= n - 1; i++) printf("%d ", min(ans[i], mx));
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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>
using namespace std;
const int sz = 5e3 + 9;
int n, m;
vector<pair<int, int>> v[sz];
int a, b, g;
int ans[sz];
int first, t;
set<pair<int, pair<int, int>>> mys;
int dfs1(int idx, int to, int val, int par = -1) {
  if (idx == to) return 1;
  int ret = 0;
  for (auto u : v[idx]) {
    int nxt = u.first;
    int i = u.second;
    if (nxt == par) continue;
    int tmp = dfs1(nxt, to, val, idx);
    if (tmp) ans[i] = val;
    ret |= tmp;
  }
  return ret;
}
int dfs2(int idx, int to, int par = -1) {
  if (idx == to) return -1;
  int ret = 1e9;
  for (auto u : v[idx]) {
    int nxt = u.first;
    int i = u.second;
    if (nxt == par) continue;
    int tmp = dfs2(nxt, to, idx);
    if (tmp != 1e9) {
      if (~tmp)
        ret = min(ret, min(tmp, ans[i]));
      else
        ret = min(ret, ans[i]);
    }
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d %d", &first, &t);
    v[first].push_back(make_pair(t, i));
    v[t].push_back(make_pair(first, i));
    ans[i] = 1e9;
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &a, &b, &g);
    mys.insert(make_pair(g, make_pair(a, b)));
  }
  for (auto u : mys) {
    g = u.first;
    a = u.second.first;
    b = u.second.second;
    dfs1(a, b, g);
  }
  for (auto u : mys) {
    g = u.first;
    a = u.second.first;
    b = u.second.second;
    int tmp = dfs2(a, b);
    if (tmp != g) {
      puts("-1");
      return 0;
    }
  }
  int mx = 1e6;
  for (int i = 1; i <= n - 1; i++) printf("%d ", min(ans[i], mx));
  return 0;
}

```