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 JAVA):

Solution:
'use strict'

const n = +readline();
const edges = Array.from({ length: n - 1 }, () => readline().split(' ').map(i => +i - 1));
const passengers = Array.from({ length: +readline() }, () => readline().split(' ').map(i => +i)).map(p => [ p[0] - 1, p[1] - 1, p[2] ]);

const rootTree = (graph, n) => {
    // Do a DFS to (arbitrarily) root the tree and record depths
    const parent = Array(n).fill(null); parent[0] = Infinity;
    const parentEdge = Array(n).fill(null); parentEdge[0] = Infinity;
    const depth = Array(n).fill(null); depth[0] = 0;
    const stack = [ 0 ];
    while (stack.length) {
        const node = stack.pop();
        Object.keys(graph[node]).forEach((k) => {
            const edgeId = graph[node][k];
            if (parent[k] !== null) return;
            parent[k] = +node;
            parentEdge[k] = edgeId;
            depth[k] = depth[node] + 1;
            stack.unshift(k);
        });
    }
    return { parent, parentEdge, depth };
}

function* getLCAEdges(parent, parentEdge, depth, source, dest) {
    // Find shortest path by walking both nodes up till their LCA
    // Returns the edge ids in arbitrary order!
    let A = source, B = dest;

    // Start them at the same depth first
    let dA = depth[A], dB = depth[B];
    const r = [];
    while (dA > dB) {
        yield parentEdge[A];
        A = parent[A];
        dA -= 1;
    }
    while (dA < dB) {
        yield parentEdge[B];
        B = parent[B];
        dB -= 1;
    }
    // Walk up both paths until find lca
    while (A !== B) {
        yield parentEdge[A];
        A = parent[A];
        yield parentEdge[B];
        B = parent[B];
    }
    return r;
}

const problem = (n, edges, passengers) => {
    const graph = {};

    edges.forEach((a, i) => {
        const x = a[0], y = a[1];
        if (!graph[x]) graph[x] = {};
        if (!graph[y]) graph[y] = {};
        graph[x][y] = i;
        graph[y][x] = i;
    })

    const rt = rootTree(graph, n)
    const parent = rt.parent;
    const parentEdge = rt.parentEdge;
    const depth = rt.depth;
    // Sort by score, high to low
    passengers.sort((a, b) => b[2] - a[2]);
    // For each passenger, set any unset edges in their path to their min score
    const scores = Array(n - 1).fill(null);

    for (let i = 0; i < passengers.length; i++) {
        const p = passengers[i];
        const a = p[0];
        const b = p[1];
        const score = p[2];
        let numEdges = 0;

        for (let edgeId of getLCAEdges(parent, parentEdge, depth, a, b)) {
            if (scores[edgeId] === null) {
                numEdges += 1;
                scores[edgeId] = score;
            } else {
                // Since sorted any existing scores must higher or equal
                // Can't overwrite lower because will lower the previous's min score
                if (scores[edgeId] === score) numEdges += 1
            }
        }
        // Need to set at least one edge
        if (numEdges == 0) return -1;
    }

    return scores.map(i => i || 1).join(' ');
}

print(problem(n, edges, passengers));