Given a tree with n nodes numbered from 1 to n. Each node i has an associated value V_i.

If the simple path from u_1 to u_m consists of m nodes namely u_1 → u_2 → u_3 → ... u_{m-1} → u_{m}, then its alternating function A(u_{1},u_{m}) is defined as A(u_{1},u_{m}) = ∑_{i=1}^{m} (-1)^{i+1} ⋅ V_{u_{i}}. A path can also have 0 edges, i.e. u_{1}=u_{m}.

Compute the sum of alternating functions of all unique simple paths. Note that the paths are directed: two paths are considered different if the starting vertices differ or the ending vertices differ. The answer may be large so compute it modulo 10^{9}+7. 

Input

The first line contains an integer n (2 ≤ n ≤ 2⋅10^{5} ) — the number of vertices in the tree.

The second line contains n space-separated integers V_1, V_2, …, V_n (-10^9≤ V_i ≤ 10^9) — values of the nodes.

The next n-1 lines each contain two space-separated integers u and v (1≤ u, v≤ n, u ≠ v) denoting an edge between vertices u and v. It is guaranteed that the given graph is a tree.

Output

Print the total sum of alternating functions of all unique simple paths modulo 10^{9}+7. 

Examples

Input

4
-4 1 5 -2
1 2
1 3
1 4


Output

40


Input

8
-2 6 -4 -4 -9 -3 -7 23
8 2
2 3
1 4
6 5
7 6
4 7
5 8


Output

4

Note

Consider the first example.

A simple path from node 1 to node 2: 1 → 2 has alternating function equal to A(1,2) = 1 ⋅ (-4)+(-1) ⋅ 1 = -5.

A simple path from node 1 to node 3: 1 → 3 has alternating function equal to A(1,3) = 1 ⋅ (-4)+(-1) ⋅ 5 = -9.

A simple path from node 2 to node 4: 2 → 1 → 4 has alternating function A(2,4) = 1 ⋅ (1)+(-1) ⋅ (-4)+1 ⋅ (-2) = 3.

A simple path from node 1 to node 1 has a single node 1, so A(1,1) = 1 ⋅ (-4) = -4.

Similarly, A(2, 1) = 5, A(3, 1) = 9, A(4, 2) = 3, A(1, 4) = -2, A(4, 1) = 2, A(2, 2) = 1, A(3, 3) = 5, A(4, 4) = -2, A(3, 4) = 7, A(4, 3) = 7, A(2, 3) = 10, A(3, 2) = 10. So the answer is (-5) + (-9) + 3 + (-4) + 5 + 9 + 3 + (-2) + 2 + 1 + 5 + (-2) + 7 + 7 + 10 + 10 = 40.

Similarly A(1,4)=-2, A(2,2)=1, A(2,1)=5, A(2,3)=10, A(3,3)=5, A(3,1)=9, A(3,2)=10, A(3,4)=7, A(4,4)=-2, A(4,1)=2, A(4,2)=3 , A(4,3)=7 which sums upto 40. 
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.util.StringTokenizer;
import java.io.Writer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author kessido
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        OutputWriter out = new OutputWriter(outputStream);
        EAlternatingTree solver = new EAlternatingTree();
        solver.solve(1, in, out);
        out.close();
    }

    static class EAlternatingTree {
        final long MOD = MathExtensions.DEFAULT_MOD;
        int[][] g;
        int[][] numberOfChildrens;
        int[][] numberOfNotChildrens;
        int[] parity;
        int[] totalPerParity;
        long[] usage;

        public void solve(int testNumber, InputReader in, OutputWriter out) {
            int n = in.NextInt();
            long[] v = in.NextLongArray(n);
            for (int i = 0; i < n; i++) {
                v[i] %= MOD;
                if (v[i] < 0)
                    v[i] += MOD;
            }

            numberOfChildrens = new int[n][2];
            numberOfNotChildrens = new int[n][2];
            parity = new int[n];
            totalPerParity = new int[n];
            usage = new long[n];
            g = GraphExtension.parseGraph_n_m_and_m_lines_of_u_v_noSelfLoops(n, n - 1, in, false, false);
            setParity(0, -1, 0);
            countChildren(0, -1);
            long res = 0;
            for (int i = 0; i < n; i++) {
                usage[i] %= MOD;
                if (usage[i] < 0) usage[i] += MOD;
            }
            for (int i = 0; i < n; i++) {
                res = (res + (usage[i] * v[i]) % MOD) % MOD;
            }
            out.println(res);
        }

        private void countChildren(int i, int p) {
            numberOfChildrens[i][0]++;
            usage[i]++;
            for (int c : g[i]) {
                if (p == c) continue;
                countChildren(c, i);
                long myCount = numberOfChildrens[i][0] + numberOfChildrens[i][1];
                long cCount = numberOfChildrens[c][0] + numberOfChildrens[c][1];

                usage[i] += numberOfChildrens[c][1] * myCount;
                usage[i] -= numberOfChildrens[c][0] * myCount;
                usage[i] += numberOfChildrens[i][0] * cCount;
                usage[i] -= numberOfChildrens[i][1] * cCount;

                for (int j = 0; j < 2; j++) {
                    numberOfChildrens[i][j] += numberOfChildrens[c][1 - j];
                }
            }
            for (int j = 0; j < 2; j++) {
                numberOfNotChildrens[i][j] = totalPerParity[parity[i] ^ j] - numberOfChildrens[i][j];
            }
            long totalOutside = numberOfNotChildrens[i][0] + numberOfNotChildrens[i][1];
            long totalInside = numberOfChildrens[i][0] + numberOfChildrens[i][1];
            usage[i] += numberOfChildrens[i][0] * totalOutside;
            usage[i] -= numberOfChildrens[i][1] * totalOutside;
            usage[i] += numberOfNotChildrens[i][0] * totalInside;
            usage[i] -= numberOfNotChildrens[i][1] * totalInside;
        }

        private void setParity(int i, int parent, int myParity) {
            parity[i] = myParity;
            totalPerParity[myParity]++;
            for (int c : g[i]) {
                if (c != parent) setParity(c, i, myParity ^ 1);
            }
        }

    }

    static class GraphExtension {
        public static int[][] parseGraph_n_e_array_of_of_u_v_noSelfLoops(final int n, final int[][] e, final boolean directed) {
            int[] edgesCount = new int[n];
            for (int[] i : e) {
                edgesCount[i[0]]++;
                if (!directed)
                    edgesCount[i[1]]++;
            }
            int[][] graph = new int[n][];
            for (int i = 0; i < n; i++) {
                graph[i] = new int[edgesCount[i]];
            }
            for (int[] i : e) {
                graph[i[0]][--edgesCount[i[0]]] = i[1];
                if (!directed)
                    graph[i[1]][--edgesCount[i[1]]] = i[0];
            }
            return graph;
        }

        public static int[][] parseGraph_n_m_and_m_lines_of_u_v_noSelfLoops(final int n, final int m, final InputReader in, final boolean zeroBasedIndex, final boolean directed) {
            int[][] e = in.NextIntMatrix(m, 2, zeroBasedIndex ? 0 : -1);
            return parseGraph_n_e_array_of_of_u_v_noSelfLoops(n, e, directed);
        }

    }

    static class OutputWriter extends PrintWriter {
        public OutputWriter(Writer out) {
            super(out);
        }

        public OutputWriter(File file) throws FileNotFoundException {
            super(file);
        }

        public OutputWriter(OutputStream out) {
            super(out);
        }

    }

    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine(), " \t\n\r\f,");
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int NextInt() {
            return Integer.parseInt(next());
        }

        public long NextLong() {
            return Long.parseLong(next());
        }

        public int[][] NextIntMatrix(int n, int m, int offset) {
            int[][] a = new int[n][m];
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    a[i][j] = NextInt() + offset;
                }
            }
            return a;
        }

        public long[] NextLongArray(int n) {
            return NextLongArray(n, 0);
        }

        public long[] NextLongArray(int n, int offset) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = NextLong() + offset;
            }
            return a;
        }

    }

    static class MathExtensions {
        public static final long DEFAULT_MOD = 1_000_000_007L;

    }
}

