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.util.List;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.util.LinkedList;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        R474E solver = new R474E();
        solver.solve(1, in, out);
        out.close();
    }

    static class R474E {
        long mod = (int) 1e9 + 7;
        int[] v;
        List<GraphUtil.Edge>[] g;
        Node[] nodes;
        long ans = 0;

        void dfs(int u) {
            Node node = nodes[u];
            node.no++;
            node.so = v[u];
            ans = (ans + v[u] + mod) % mod;
            for (GraphUtil.Edge e : g[u]) {
                if (nodes[e.to] == null) {
                    Node child = nodes[e.to] = new Node(e.to, u);
                    dfs(child.id);
                    ans += (node.so + node.se) % mod * (child.ne + child.no) % mod;// start left, left part
                    ans += (child.so - child.se) % mod * (node.ne - node.no) % mod;// right part

                    ans += (child.so + child.se) % mod * (node.ne + node.no) % mod;// start right, right part
                    ans += (node.so - node.se) % mod * (child.ne - child.no) % mod;// left part

                    node.so = (node.so + child.se) % mod;
                    node.se = (node.se + child.so) % mod;
                    node.no = (node.no + child.ne) % mod;
                    node.ne = (node.ne + child.no) % mod;
                    node.se = (node.se - v[u] * child.no % mod + mod) % mod;
                    node.so = (node.so + v[u] * child.ne % mod + mod) % mod;
                    ans = (ans % mod + mod) % mod;
                }
            }
        }

        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            int n = in.ni();
            v = in.na(n);
            g = in.readEdgesGraph(n, n - 1, true);
            nodes = new Node[n];
            nodes[0] = new Node(0, -1);
            dfs(0);
        /*for (int i = 0; i < n; i++) {
            used = new boolean[n];
            used[i] = true;
            dfs1(i, true);
        }
        System.out.println("ans2 = " + ans2);
        */
            out.println((ans % mod + mod) % mod);
        }

        class Node {
            int id;
            int p;
            long no;
            long ne;
            long so;
            long se;
            List<Integer> children;

            public Node(int id, int p) {
                this.id = id;
                this.p = p;
                children = new LinkedList<>();
            }

        }

    }

    static class GraphUtil {
        public static class Edge {
            public int from;
            public int to;
            public int id;
            public long w;

            public Edge(int from, int to) {
                this.from = from;
                this.to = to;
            }

            public Edge(int from, int to, int id) {
                this.from = from;
                this.to = to;
                this.id = id;
            }

            public Edge(int from, int to, int id, long w) {
                this.from = from;
                this.to = to;
                this.id = id;
                this.w = w;
            }

        }

    }

    static class FastScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

        public String ns() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    String rl = in.readLine();
                    if (rl == null) {
                        return null;
                    }
                    st = new StringTokenizer(rl);
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int ni() {
            return Integer.parseInt(ns());
        }

        public int[] na(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = ni();
            return a;
        }

        public List<GraphUtil.Edge>[] readEdgesGraph(int n, int m, boolean bidirected) {
            List<GraphUtil.Edge>[] g = new List[n];
            for (int i = 0; i < n; i++) {
                g[i] = new LinkedList<>();
            }
            for (int i = 0; i < m; i++) {
                int u = ni() - 1;
                int v = ni() - 1;
                g[u].add(new GraphUtil.Edge(u, v, i));
                if (bidirected)
                    g[v].add(new GraphUtil.Edge(v, u, i));
            }

            return g;
        }

    }
}

