#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e3 + 10, INF = 0x3f3f3f3f;
struct edge {
  int x;
  int y;
} e[maxn];
int w[maxn][maxn];
int num[maxn];
vector<int> vec[maxn];
int depth[maxn];
int fa[maxn];
void dfs(int cur, int p, int dep) {
  fa[cur] = p;
  depth[cur] = dep;
  for (int x : vec[cur]) {
    if (x != p) {
      dfs(x, cur, dep + 1);
    }
  }
  num[cur] = INF;
}
bool lca_modify(int x, int y, int k) {
  int minn = INF;
  if (depth[x] < depth[y]) swap(x, y);
  while (depth[x] != depth[y]) {
    if (num[x] == INF) {
      num[x] = k;
      w[x][fa[x]] = w[fa[x]][x] = k;
    }
    minn = min(minn, num[x]);
    x = fa[x];
  }
  while (x != y) {
    if (num[x] == INF) {
      num[x] = k;
      w[x][fa[x]] = w[fa[x]][x] = k;
    }
    minn = min(minn, num[x]);
    x = fa[x];
    if (num[y] == INF) {
      num[y] = k;
      w[y][fa[y]] = w[fa[y]][y] = k;
    }
    minn = min(minn, num[y]);
    y = fa[y];
  }
  return minn == k;
}
const int maxm = 5e3 + 10;
struct query {
  int x, y, len;
} q[maxm];
bool cmp(query a, query b) { return a.len > b.len; }
int main() {
  int n, m;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    e[i].x = x, e[i].y = y;
    vec[x].push_back(y);
    vec[y].push_back(x);
  }
  dfs(1, 0, 1);
  cin >> m;
  int x, y, len;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y >> len;
    q[i].x = x, q[i].y = y, q[i].len = len;
  }
  sort(q + 1, q + 1 + m, cmp);
  int flag = 1;
  for (int i = 1; i <= m; i++) {
    x = q[i].x, y = q[i].y, len = q[i].len;
    if (lca_modify(x, y, len) == false) {
      flag = 0;
      break;
    }
  }
  if (!flag) {
    cout << -1 << endl;
  } else {
    for (int i = 0; i < n - 1; i++) {
      x = e[i].x, y = e[i].y;
      if (x > y) swap(x, y);
      if (w[x][y] == 0)
        cout << 1000000;
      else
        cout << w[x][y];
      if (i != n - 2)
        cout << " ";
      else
        cout << endl;
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e3 + 10, INF = 0x3f3f3f3f;
struct edge {
  int x;
  int y;
} e[maxn];
int w[maxn][maxn];
int num[maxn];
vector<int> vec[maxn];
int depth[maxn];
int fa[maxn];
void dfs(int cur, int p, int dep) {
  fa[cur] = p;
  depth[cur] = dep;
  for (int x : vec[cur]) {
    if (x != p) {
      dfs(x, cur, dep + 1);
    }
  }
  num[cur] = INF;
}
bool lca_modify(int x, int y, int k) {
  int minn = INF;
  if (depth[x] < depth[y]) swap(x, y);
  while (depth[x] != depth[y]) {
    if (num[x] == INF) {
      num[x] = k;
      w[x][fa[x]] = w[fa[x]][x] = k;
    }
    minn = min(minn, num[x]);
    x = fa[x];
  }
  while (x != y) {
    if (num[x] == INF) {
      num[x] = k;
      w[x][fa[x]] = w[fa[x]][x] = k;
    }
    minn = min(minn, num[x]);
    x = fa[x];
    if (num[y] == INF) {
      num[y] = k;
      w[y][fa[y]] = w[fa[y]][y] = k;
    }
    minn = min(minn, num[y]);
    y = fa[y];
  }
  return minn == k;
}
const int maxm = 5e3 + 10;
struct query {
  int x, y, len;
} q[maxm];
bool cmp(query a, query b) { return a.len > b.len; }
int main() {
  int n, m;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    e[i].x = x, e[i].y = y;
    vec[x].push_back(y);
    vec[y].push_back(x);
  }
  dfs(1, 0, 1);
  cin >> m;
  int x, y, len;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y >> len;
    q[i].x = x, q[i].y = y, q[i].len = len;
  }
  sort(q + 1, q + 1 + m, cmp);
  int flag = 1;
  for (int i = 1; i <= m; i++) {
    x = q[i].x, y = q[i].y, len = q[i].len;
    if (lca_modify(x, y, len) == false) {
      flag = 0;
      break;
    }
  }
  if (!flag) {
    cout << -1 << endl;
  } else {
    for (int i = 0; i < n - 1; i++) {
      x = e[i].x, y = e[i].y;
      if (x > y) swap(x, y);
      if (w[x][y] == 0)
        cout << 1000000;
      else
        cout << w[x][y];
      if (i != n - 2)
        cout << " ";
      else
        cout << endl;
    }
  }
  return 0;
}

```