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 = 2e5 + 7;
const int M = 16;
const int mod = 1e9 + 7;
int n, q;
int cnt;
int pp[N];
int dep[N];
int ans[N];
int tin[N];
int tout[N];
int up[N][M];
int mn[N][M];
pair<int, pair<int, int> > t[N];
vector<int> v[N];
vector<pair<int, int> > g[N];
vector<pair<int, int> > k;
map<pair<int, int>, int> m;
void dfs(int x, int p, int z) {
  if (x == p) return;
  int xx = m[{x, pp[x]}];
  ans[xx] = max(ans[xx], z);
  dfs(pp[x], p, z);
}
void dfs2(int x, int p) {
  up[x][0] = p;
  tin[x] = ++cnt;
  dep[x] = dep[p] + 1;
  pp[x] = p;
  for (int i = 1; i < M; i++) {
    up[x][i] = up[up[x][i - 1]][i - 1];
  }
  for (auto y : v[x]) {
    if (y != p) {
      dfs2(y, x);
    }
  }
  tout[x] = cnt;
}
void dfs1(int x, int p, int z) {
  mn[x][0] = z;
  up[x][0] = p;
  tin[x] = ++cnt;
  dep[x] = dep[p] + 1;
  pp[x] = p;
  for (int i = 1; i < M; i++) {
    up[x][i] = up[up[x][i - 1]][i - 1];
    mn[x][i] = min(mn[x][i - 1], mn[up[x][i - 1]][i - 1]);
  }
  for (auto y : g[x]) {
    if (y.first != p) {
      dfs1(y.first, x, y.second);
    }
  }
  tout[x] = cnt;
}
bool upper(int x, int y) { return (tin[x] <= tin[y] && tout[x] >= tout[y]); }
int lca(int x, int y) {
  if (upper(x, y)) return x;
  if (upper(y, x)) return y;
  for (int i = M - 1; i >= 0; i--) {
    if (!upper(up[x][i], y)) {
      x = up[x][i];
    }
  }
  return up[x][0];
}
int get(int x, int z) {
  int res = 1e9;
  for (int i = M - 1; i >= 0; i--) {
    if ((1 << i) <= z) {
      z -= (1 << i);
      res = min(res, mn[x][i]);
      x = up[x][i];
    }
  }
  return res;
}
void solve1() {
  cin >> n;
  vector<pair<int, int> > fg;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    m[{x, y}] = m[{y, x}] = i;
    v[x].push_back(y);
    v[y].push_back(x);
    fg.push_back({x, y});
  }
  cin >> q;
  dfs2(1, 1);
  for (int i = 1; i <= q; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    int zz = lca(x, y);
    dfs(x, zz, z);
    dfs(y, zz, z);
    t[i] = {x, {y, z}};
  }
  for (int i = 1; i < n; i++) {
    if (!ans[i]) ans[i] = 1e6;
  }
  for (int i = 1; i < n; i++) {
    g[fg[i - 1].first].push_back(
        {fg[i - 1].second, ans[m[{fg[i - 1].first, fg[i - 1].second}]]});
    g[fg[i - 1].second].push_back(
        {fg[i - 1].first, ans[m[{fg[i - 1].first, fg[i - 1].second}]]});
  }
  cnt = 0;
  dep[1] = 0;
  dfs1(1, 1, 1e9);
  for (int i = 1; i <= q; i++) {
    int x = lca(t[i].first, t[i].second.first);
    int z = min(get(t[i].first, dep[t[i].first] - dep[x]),
                get(t[i].second.first, dep[t[i].second.first] - dep[x]));
    if (z != t[i].second.second) {
      cout << -1;
      return;
    }
  }
  for (int i = 1; i < n; i++) {
    cout << ans[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  int cghf = 1;
  while (cghf--) {
    solve1();
  }
}