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 = 5003;
int n, m, head[N], tot, ans[N], id[N];
int sz[N], dep[N], fa[N], top[N], son[N];
struct Edge {
  int to, nxt;
} e[N * 2];
struct SEdge {
  int x, y;
} se[N];
struct node {
  int x, y, w;
} q[N];
void addEdge(int x, int y) {
  e[++tot] = (Edge){y, head[x]};
  head[x] = tot;
  e[++tot] = (Edge){x, head[y]};
  head[y] = tot;
}
void dfs1(int x, int fz) {
  fa[x] = fz;
  sz[x] = 1;
  dep[x] = dep[fz] + 1;
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == fz) continue;
    dfs1(y, x);
    sz[x] += sz[y];
    if (sz[y] > sz[son[x]]) son[x] = y;
  }
}
void dfs2(int x, int t) {
  top[x] = t;
  if (son[x]) dfs2(son[x], t);
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == fa[x] || y == son[x]) continue;
    dfs2(y, y);
  }
}
int lca(int x, int y) {
  while (top[x] ^ top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    x = fa[top[x]];
  }
  return dep[x] < dep[y] ? x : y;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    addEdge(x, y);
    se[i] = (SEdge){x, y};
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d %d %d", &q[i].x, &q[i].y, &q[i].w);
  dfs1(1, 0);
  dfs2(1, 1);
  for (int i = 1; i < n; i++) {
    int &x = se[i].x, &y = se[i].y;
    if (fa[x] != y) swap(x, y);
    id[x] = i;
  }
  for (int i = 1; i <= m; i++) {
    int x = q[i].x, y = q[i].y, w = q[i].w, l = lca(x, y);
    while (x != l) {
      ans[id[x]] = max(ans[id[x]], w);
      x = fa[x];
    }
    while (y != l) {
      ans[id[y]] = max(ans[id[y]], w);
      y = fa[y];
    }
  }
  for (int i = 1; i <= m; i++) {
    int x = q[i].x, y = q[i].y, w = q[i].w, l = lca(x, y), mn = 1e9;
    while (x != l) mn = min(mn, ans[id[x]]), x = fa[x];
    while (y != l) mn = min(mn, ans[id[y]]), y = fa[y];
    if (mn != w) printf("-1\n"), exit(0);
  }
  for (int i = 1; i < n; i++) printf("%d ", ans[i] == 0 ? 1 : ans[i]);
  return 0;
}