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 mod = 1e9 + 7;
const int N = 1e4 + 5;
int head[N], fa[N], deep[N], pos[N];
int tot = 1, val[N];
struct node {
  int v, nxt, id;
} edge[N];
struct Node {
  int u, v, w;
} e[N];
bool cmp(Node a, Node b) { return a.w > b.w; }
void add(int u, int v, int i) {
  edge[tot].v = v;
  edge[tot].id = i;
  edge[tot].nxt = head[u];
  head[u] = tot++;
}
void dfs(int u, int fat) {
  for (int i = head[u]; i != -1; i = edge[i].nxt) {
    int v = edge[i].v;
    if (v == fat) continue;
    pos[edge[i].id] = v;
    fa[v] = u;
    deep[v] = deep[u] + 1;
    dfs(v, u);
  }
}
int main() {
  int n, m;
  scanf("%d", &n);
  memset(head, -1, sizeof head);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v, i);
    add(v, u, i);
  }
  deep[1] = 1;
  dfs(1, 0);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
  }
  sort(e + 1, e + 1 + m, cmp);
  for (int i = 1; i <= m; i++) {
    int u = e[i].u, v = e[i].v, w = e[i].w;
    bool flag = false;
    if (deep[u] < deep[v]) swap(u, v);
    while (deep[u] != deep[v]) {
      if ((!val[u]) || val[u] == w) {
        val[u] = w;
        flag = true;
      }
      u = fa[u];
    }
    while (u != v) {
      if ((!val[u]) || val[u] == w) {
        val[u] = w;
        flag = true;
      }
      if ((!val[v]) || val[v] == w) {
        val[v] = w;
        flag = true;
      }
      u = fa[u], v = fa[v];
    }
    if (!flag) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    if (val[pos[i]]) {
      printf("%d ", val[pos[i]]);
    } else {
      printf("1000000 ");
    }
  }
  return 0;
}