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.util.ArrayList;
import java.util.HashMap;
import java.util.Scanner;
 
public class P960E {
	private static Scanner in;

    private static int n;
    private static int[] a;
    private static HashMap<Integer, ArrayList<Integer>> g;

    private static int[] size;
    private static long[] sumA;

    private static long ans = 0;
    
    private static int MOD = (int)1e9+7;


	public static void main(String[] args) {
		in = new Scanner(System.in);
		n = in.nextInt();

        g = new HashMap<>();
        for(int i=0; i<n; i++) {
            g.put(i, new ArrayList<Integer>());
        }

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

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

            g.get(p).add(q);
            g.get(q).add(p);
        }

        size = new int[n];
        sumA = new long[n];

        dfs(0, -1);
        dfs2(0, -1);

        System.out.println((ans+MOD)%MOD);

	}

    public static void dfs(int u, int p) {

    	size[u] = 1;
    	sumA[u] = a[u];
    	
        for(int v: g.get(u)) {
            if (v==p) continue;

            dfs(v, u);

            size[u] += size[v];

            sumA[u] += size[v]*1L*a[u]-sumA[v];
        }
    }

    public static void dfs2(int u, int p) {
        //u is at the root
        ans = (ans+sumA[u])%MOD;

        for(int v: g.get(u)) {
            if (v==p) continue;

            size[u] -= size[v];
            sumA[u] -= (size[v]*1L*a[u]-sumA[v]);
            size[v] += size[u];
            sumA[v] += (size[u]*1L*a[v]-sumA[u]);

            dfs2(v, u);
            
            sumA[v] -= (size[u]*1L*a[v]-sumA[u]);
            size[v] -= size[u];
            sumA[u] += (size[v]*1L*a[u]-sumA[v]);
            size[u] += size[v];

        }
    }
   
 
}