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 power(long long x, long long y, long long z) {
  long long ret = 1;
  while (y > 0) {
    if (y & 1) ret = (ret * x) % z;
    x = (x * x) % z;
    y >>= 1;
  }
  return ret;
}
const long long N = 5e3 + 5, LG = 20, MX = 1e6;
long long dp[N][LG], L[N], val[N][N];
vector<long long> V[N];
pair<long long, long long> E[N];
void dfs(long long x, long long p) {
  L[x] = L[p] + 1;
  dp[x][0] = p;
  for (auto u : V[x]) {
    if (u == p) continue;
    dfs(u, x);
  }
  return;
}
long long LCA(long long a, long long b) {
  if (L[a] > L[b]) swap(a, b);
  for (long long j = LG - 1; j >= 0; --j) {
    if (L[b] - (1LL << j) >= L[a]) b = dp[b][j];
  }
  if (a == b) return a;
  for (long long j = LG - 1; j >= 0; --j) {
    if (dp[a][j] != dp[b][j]) {
      a = dp[a][j];
      b = dp[b][j];
    }
  }
  return dp[a][0];
}
struct query {
  long long a, b, c;
} Q[N];
bool cmp(struct query X, struct query Y) { return X.c > Y.c; }
int main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 1; i <= n - 1; ++i) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    E[i] = make_pair(x, y);
    V[x].push_back(y);
    V[y].push_back(x);
  }
  dfs(1, 0);
  for (long long j = 1; j <= LG - 1; ++j)
    for (long long i = 1; i <= n; ++i) dp[i][j] = dp[dp[i][j - 1]][j - 1];
  long long m;
  scanf("%lld", &m);
  bool fl = 1;
  for (long long i = 1; i <= m; ++i) {
    scanf("%lld%lld%lld", &Q[i].a, &Q[i].b, &Q[i].c);
  }
  sort(Q + 1, Q + 1 + m, cmp);
  for (long long i = 1; i <= m; ++i) {
    long long x = Q[i].a, y = Q[i].b, z = Q[i].c;
    long long lca = LCA(x, y);
    long long tmp = x;
    while (tmp != lca) {
      long long p = tmp, q = dp[tmp][0];
      if (!val[p][q]) {
        val[p][q] = z;
        val[q][p] = z;
      }
      tmp = dp[tmp][0];
    }
    tmp = y;
    while (tmp != lca) {
      long long p = tmp, q = dp[tmp][0];
      if (!val[p][q]) {
        val[p][q] = z;
        val[q][p] = z;
      }
      tmp = dp[tmp][0];
    }
  }
  for (long long i = 1; i <= m; ++i) {
    scanf("%lld%lld%lld", &Q[i].a, &Q[i].b, &Q[i].c);
    long long x = Q[i].a, y = Q[i].b, z = Q[i].c;
    long long lca = LCA(x, y);
    long long tmp = x, mn = MX;
    while (tmp != lca) {
      mn = min(mn, val[tmp][dp[tmp][0]]);
      tmp = dp[tmp][0];
    }
    tmp = y;
    while (tmp != lca) {
      mn = min(mn, val[tmp][dp[tmp][0]]);
      tmp = dp[tmp][0];
    }
    if (mn != z) fl = 0;
  }
  if (!fl)
    printf("-1\n");
  else {
    for (long long i = 1; i <= n - 1; ++i) {
      long long x = E[i].first, y = E[i].second;
      if (!val[x][y])
        printf("%lld ", MX);
      else
        printf("%lld ", val[x][y]);
    }
    printf("\n");
  }
  return 0;
}