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 = 100005;
struct edge {
  int u, v, i;
  int nxt;
} e[maxn];
struct AK {
  int u, v, w;
  bool operator<(const AK& a) const { return w > a.w; }
} ak[maxn];
int cnt;
int head[maxn];
void add_edge(int u, int v, int pos) {
  e[cnt].u = u, e[cnt].v = v, e[cnt].i = pos;
  e[cnt].nxt = head[u];
  head[u] = cnt++;
}
int val[maxn];
vector<int> vec[maxn];
bool dfs(int u, int fa, int id) {
  if (u == ak[id].v) return true;
  for (int i = head[u]; i != -1; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa) continue;
    int ee = e[i].i;
    if (dfs(v, u, id)) {
      val[ee] = max(val[ee], ak[id].w);
      vec[id].push_back(ee);
      return true;
    }
  }
  return false;
}
int main() {
  int n;
  memset(val, 0, sizeof val);
  memset(head, -1, sizeof head);
  cnt = 0;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add_edge(u, v, i);
    add_edge(v, u, i);
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int a, b, c;
    scanf("%d%d%d", &ak[i].u, &ak[i].v, &ak[i].w);
  }
  sort(ak, ak + m);
  for (int i = 0; i < m; i++) {
    int a, b, c;
    a = ak[i].u, b = ak[i].v;
    dfs(a, -1, i);
  }
  const int INF = 0x3f3f3f3f;
  for (int i = 0; i < m; i++) {
    int minv = INF;
    for (int j : vec[i]) {
      if (val[j] < minv) {
        minv = val[j];
      }
    }
    if (minv != ak[i].w) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    if (i != 1) cout << " ";
    if (val[i] == 0)
      cout << 1;
    else
      cout << val[i];
  }
}