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.BufferedReader;
// import java.io.FileInputStream;
// import java.io.FileOutputStream;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
import java.util.Map.Entry;

import static java.lang.Math.*;
import static java.util.Arrays.copyOf;
import static java.util.Arrays.fill;
import static java.util.Arrays.sort;
import static java.util.Collections.reverseOrder;
import static java.util.Collections.sort;

public class Main {
    FastScanner in;
    PrintWriter out;

    int n;
    long mod = (long) 1e9 + 7;
    long[] v;
    ArrayList<Integer>[] graph;
    long[][] down, up, downCnt, upCnt;
    long ans;

    private void solve() throws IOException {
        n = in.nextInt();
        up = new long[n][2];
        down = new long[n][2];
        upCnt = new long[n][2];
        downCnt = new long[n][2];
        v = new long[n];
        for (int i = 0; i < n; i++) {
            v[i] = in.nextLong();
        }

        graph = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            graph[i] = new ArrayList<>();
        }
        for (int i = 1, v, u; i < n; i++) {
            v = in.nextInt() - 1;
            u = in.nextInt() - 1;
            graph[v].add(u);
            graph[u].add(v);
        }

        ans = 0;
        dfs(0, -1);
        for (int i = 0; i < n; i++) {
            ans -= v[i];
            ans %= mod;
        }
        out.print((ans + mod) % mod);
    }

    private void dfs(int v, int p) {
        if (p != -1) {
            up[v][0] -= up[p][1] + down[p][1];
            upCnt[v][0] += upCnt[p][1] + downCnt[p][1];
            up[v][0] += this.v[v] * upCnt[v][0] % mod;
            up[v][0] %= mod;

            up[v][1] -= up[p][0] + down[p][0];
            upCnt[v][1] += upCnt[p][0] + downCnt[p][0];
            up[v][1] += this.v[v] * upCnt[v][1] % mod;
            up[v][1] %= mod;

        }

        downCnt[v][1]++;
        down[v][1] = this.v[v];

        ans += 2 * (up[v][1] + down[v][1]);
        ans %= mod;

        for (int u : graph[v])
            if (u != p) {
                dfs(u, v);

                down[v][0] -= down[u][1];
                downCnt[v][0] += downCnt[u][1];
                down[v][0] += this.v[v] * downCnt[u][1] % mod;
                down[v][0] %= mod;

                down[v][1] -= down[u][0];
                downCnt[v][1] += downCnt[u][0];
                down[v][1] += this.v[v] * downCnt[u][0] % mod;
                down[v][1] %= mod;
            }
    }

    class FastScanner {
        StringTokenizer st;
        BufferedReader br;

        FastScanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        boolean hasNext() throws IOException {
            return br.ready() || (st != null && st.hasMoreTokens());
        }

        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        double nextDouble() throws IOException {
            return Double.parseDouble(next().replace(',', '.'));
        }

        String nextLine() throws IOException {
            return br.readLine();
        }

        boolean hasNextLine() throws IOException {
            return br.ready();
        }

    }

    private void run() throws IOException {
        in = new FastScanner(System.in); // new FastScanner(new FileInputStream(".in"));
        out = new PrintWriter(System.out); // new PrintWriter(new FileOutputStream(".out"));

        solve();

        out.flush();
        out.close();
    }

    public static void main(String[] args) throws IOException {
        new Main().run();
    }
}
