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;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long fpow(long long x, long long y, long long mod) {
  long long ans = 1ll;
  for (; y; y >>= 1) {
    if (y & 1) ans = ans * x % mod;
    x = x * x % mod;
  }
  return ans;
}
const int maxd = 2e5 + 10;
vector<pair<long long, long long> > G[maxd];
long long d[maxd];
long long a[maxd], c[3][maxd], pos[maxd], fa[maxd];
void dfs(long long x) {
  for (auto p : G[x]) {
    long long v = p.first;
    if (!d[v]) {
      d[v] = d[x] + 1;
      fa[v] = x;
      pos[v] = p.second;
      dfs(v);
    }
  }
}
void change(long long x, long long y, long long z) {
  if (d[x] < d[y]) swap(x, y);
  while (d[x] > d[y]) {
    a[pos[x]] = max(z, a[pos[x]]);
    x = fa[x];
  }
  while (x != y) {
    a[pos[x]] = max(z, a[pos[x]]);
    x = fa[x];
    a[pos[y]] = max(z, a[pos[y]]);
    y = fa[y];
  }
}
bool check(long long x, long long y, long long z) {
  long long now = 1e9;
  if (d[x] < d[y]) swap(x, y);
  while (d[x] > d[y]) {
    now = min(a[pos[x]], now);
    x = fa[x];
  }
  while (x != y) {
    now = min(a[pos[x]], now);
    x = fa[x];
    now = min(a[pos[y]], now);
    y = fa[y];
  }
  return now == z;
}
int main() {
  long long n;
  scanf("%lld", &n);
  long long x, y;
  for (register long long i = 1; i <= n - 1; i++) {
    scanf("%lld %lld", &x, &y);
    G[x].push_back(pair<long long, long long>(y, i));
    G[y].push_back(pair<long long, long long>(x, i));
  }
  d[1] = 1;
  dfs(1);
  long long q;
  scanf("%lld", &q);
  for (register long long i = 1; i <= q; i++) {
    scanf("%lld %lld %lld", &c[0][i], &c[1][i], &c[2][i]);
    change(c[0][i], c[1][i], c[2][i]);
  }
  long long flag = 1;
  for (register long long i = 1; i <= q; i++)
    flag &= check(c[0][i], c[1][i], c[2][i]);
  if (flag)
    for (register long long i = 1; i <= n - 1; i++)
      printf("%lld ", max(a[i], 1ll));
  else
    printf("%s\n", "-1");
  return 0;
}