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.awt.*;
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintStream;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.Arrays;
import java.util.Scanner;

public class CF_960E {
    static int n;
    static int[] v;
    static ArrayList<ArrayList<Integer>> adj;

    static long MOD = (int) 1e9 + 7;

    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        PrintStream out = new PrintStream(new BufferedOutputStream(System.out));

        n = in.nextInt();

        v = new int[n];
        for (int i = 0; i < n; i++) v[i] = in.nextInt();

        adj = new ArrayList<>();
        sizes = new int[n];
        ps = new int[n];
        for (int i = 0; i < n; i++) adj.add(new ArrayList<>());

        for (int i = 0; i < n - 1; i++) {
            int u = in.nextInt() - 1;
            int v = in.nextInt() - 1;

            adj.get(u).add(v);
            adj.get(v).add(u);
        }
        dfsSize(0, -1);

        long base = bash(0, -1, 0, 0);
        out.println(ans(0, -1, base));

        out.close();
        in.close();
    }

    static long ans(int curr, int p, long ans){
        ans = mod(ans);

        long ret = ans;
        for(Integer u: adj.get(curr)){
            if(u != p){
                ret += ans(u, curr, -(ans - v[curr] * size(curr, u)) + v[u] * size(u, curr));
                ret = mod(ret);
            }
        }
        return ret;
    }

    static int[] sizes;
    static int[] ps;
    static int dfsSize(int curr, int p){
        int ret = 1;
        for(int u: adj.get(curr)){
            if(u != p)ret += dfsSize(u, curr);
        }
        sizes[curr] = ret;
        ps[curr] = p;
        return ret;
    }
    static long size(int u, int v) {
        if(u == ps[v]) return sizes[v];
        return n - sizes[u];
    }

    static long bash(int curr, int p, long ans, int n) {
        ans = (n % 2 == 0 ? 1 : -1) * v[curr] + ans;
        ans = mod(ans);

        long ret = ans;
        for (Integer v : adj.get(curr)) {
            if (v != p) {
                ret += bash(v, curr, ans, n + 1);
                ret = mod(ret);
            }
        }
        return ret;
    }

    static long mod(long a){
        return (a + MOD) % MOD;
    }
}
