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 = (int)5e3 + 10, INF = (int)1e6;
struct node {
  int to, nxt;
} e[N << 1];
struct Info {
  int u, v, w;
  bool friend operator<(const Info& a, const Info& b) { return a.w > b.w; }
} info[N];
map<pair<int, int>, int> mp;
pair<int, int> pii;
vector<pair<int, int> > vii;
int n, m, u, v, w, tot, tim;
int head[N], dfn[N], fa[N];
inline void add(int u, int v) {
  e[tot].to = v;
  e[tot].nxt = head[u];
  head[u] = tot++;
  e[tot].to = u;
  e[tot].nxt = head[v];
  head[v] = tot++;
}
void dfs_dfn(int now, int pre) {
  dfn[now] = ++tim;
  for (int o = head[now]; ~o; o = e[o].nxt) {
    if (e[o].to == pre) continue;
    dfs_dfn(e[o].to, now);
  }
}
void dfs_mp(int now, int pre) {
  for (int o = head[now]; ~o; o = e[o].nxt) {
    int to = e[o].to;
    if (to == pre) continue;
    fa[dfn[to]] = dfn[now];
    mp[make_pair(dfn[now], dfn[to])] = INF;
    dfs_mp(e[o].to, now);
  }
}
void show() {
  for (int i = 1; i <= n; ++i) printf("u = %d dfn = %d\n", i, dfn[i]);
  for (int i = 1; i <= m; ++i) cout << info[i].w << endl;
  for (int i = 1; i <= n; ++i) printf("u = %d fa = %d\n", i, fa[dfn[i]]);
}
bool lca(int u, int v, int w) {
  int u_dfn = dfn[u], v_dfn = dfn[v], tmp_w;
  bool ok = 0;
  while (u_dfn != v_dfn) {
    while (u_dfn > v_dfn) {
      pii.first = fa[u_dfn];
      pii.second = u_dfn;
      tmp_w = mp[pii];
      if (tmp_w == INF || tmp_w == w) {
        ok = 1;
        mp[pii] = w;
      }
      u_dfn = pii.first;
    }
    while (u_dfn < v_dfn) {
      pii.first = fa[v_dfn];
      pii.second = v_dfn;
      tmp_w = mp[pii];
      if (tmp_w == INF || tmp_w == w) {
        ok = 1;
        mp[pii] = w;
      }
      v_dfn = pii.first;
    }
  }
  return ok;
}
void solve() {
  for (int i = 1; i <= n; ++i) head[i] = -1;
  tot = 0;
  for (int i = 1; i <= n - 1; ++i) {
    scanf("%d%d", &u, &v);
    vii.push_back(make_pair(u, v));
    add(u, v);
  }
  dfs_dfn(1, 0);
  fa[1] = 1;
  dfs_mp(1, 0);
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &info[i].u, &info[i].v, &info[i].w);
  }
  sort(info + 1, info + 1 + m);
  for (int i = 1; i <= m; ++i) {
    if (lca(info[i].u, info[i].v, info[i].w)) continue;
    printf("-1\n");
    return;
  }
  int l = vii.size();
  int dfn1, dfn2;
  for (int i = 0; i < l; ++i) {
    dfn1 = dfn[vii[i].first];
    dfn2 = dfn[vii[i].second];
    if (dfn1 > dfn2) swap(dfn1, dfn2);
    pii.first = dfn1;
    pii.second = dfn2;
    printf("%d ", mp[pii]);
  }
  cout << endl;
}
int main() {
  scanf("%d", &n);
  solve();
  return 0;
}