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 N = 5010;
struct edge {
  int to, nxt;
} e[N << 1];
struct node {
  int x, y, z;
} s[N];
bool cmp(node a, node b) { return a.z < b.z; }
int h[N], tot = 1, fa[N], dep[N], n, m, ln[N], X[N], Y[N], g[N][N];
void adde(int x, int y) {
  e[tot] = (edge){y, h[x]};
  h[x] = tot++;
}
void dfs(int u, int ff) {
  fa[u] = ff;
  for (int i = h[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == ff) continue;
    dep[v] = dep[u] + 1;
    dfs(v, u);
  }
}
void ins(int x, int y, int z) {
  if (dep[x] < dep[y]) swap(x, y);
  while (dep[x] > dep[y]) ln[x] = z, x = fa[x];
  while (x ^ y) ln[x] = ln[y] = z, x = fa[x], y = fa[y];
}
int qry(int x, int y) {
  int ret = 1e9;
  if (dep[x] < dep[y]) swap(x, y);
  while (dep[x] > dep[y]) ret = min(ret, ln[x]), x = fa[x];
  while (x ^ y) ret = min(ret, min(ln[x], ln[y])), x = fa[x], y = fa[y];
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y), adde(x, y), adde(y, x);
    X[i] = x, Y[i] = y;
  }
  dfs(1, 0);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d%d%d", &s[i].x, &s[i].y, &s[i].z);
  sort(s + 1, s + m + 1, cmp);
  for (int i = 2; i <= n; i++) ln[i] = 1e6;
  for (int i = 1; i <= m; i++) ins(s[i].x, s[i].y, s[i].z);
  for (int i = 1; i <= m; i++)
    if (qry(s[i].x, s[i].y) != s[i].z) return printf("-1\n"), 0;
  for (int i = 1; i <= n; i++)
    for (int j = h[i]; j; j = e[j].nxt) {
      int v = e[j].to;
      if (fa[v] ^ i) continue;
      g[i][v] = g[v][i] = ln[v];
    }
  for (int i = 1; i < n; i++) printf("%d ", g[X[i]][Y[i]]);
  printf("\n");
}