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.Scanner;

public class Main {


    static int n;
    static int[] value;
    static int[] a;
    static int[] b;
    static int[] c;

    static long[] odd;
    static long[] even;
    static long[] oddPathCount;
    static long[] evenPathCount;

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

        n = in.nextInt();
        value = new int[n + 5];
        a = new int[2 * n + 5];
        b = new int[2 * n + 5];
        c = new int[2 * n + 5];
        even = new long[n + 5];
        odd = new long[ n + 5];
        evenPathCount = new long[n + 5];
        oddPathCount = new long[n + 5];

        for (int i = 1; i <= n; ++i) {
            value[i] = in.nextInt();
        }

        int x, y;
        for (int i = 1; i < n; ++i) {
            x = in.nextInt();
            y = in.nextInt();
            a[2 * i - 1] = y;
            b[2 * i - 1] = c[x];
            c[x] = 2 * i - 1;
            a[2 * i] = x;
            b[2 * i] = c[y];
            c[y] = 2 * i;
        }

        dfs(1, 0);
        System.out.println((getAmountDFS(1, 0) + 1000000007) % 1000000007);
//        for (int i = 1; i <= n; ++i) {
//            System.out.println(even[i] + odd[i]);
//        }
    }

    public static void dfs(int index, int from) {
        int i = c[index];
        while (i != 0) {
            if (a[i] != from) {
                dfs(a[i], index);
                even[index] = even[index] + odd[a[i]] - oddPathCount[a[i]] * value[index];
                odd[index] = odd[index] + even[a[i]] + evenPathCount[a[i]] * value[index];
                evenPathCount[index] += oddPathCount[a[i]];
                oddPathCount[index] += evenPathCount[a[i]];
            }
            i = b[i];
        }
        oddPathCount[index] += 1;
        odd[index] = odd[index] + value[index];
    }


    public static long getAmountDFS(int index, int from) {
        long amount = 0;
        long temp;
        int i = c[index];
        while (i != 0) {
            if (a[i] != from) {
                temp = getAmountDFS(a[i], index);
                temp = (temp + even[a[i]] * (oddPathCount[index] - evenPathCount[a[i]] - 1)) %  1000000007;
                temp = (temp + evenPathCount[a[i]] * (odd[index] - even[a[i]] - evenPathCount[a[i]] * value[index] - value[index])) % 1000000007;
                temp = (temp + 2 * (even[a[i]] + value[index] * evenPathCount[a[i]])) % 1000000007;
//                amount = (amount + getAmountDFS(a[i], index)) % 1000000007;
//                amount = (amount +
//                        even[a[i]] * (oddPathCount[index] - evenPathCount[a[i]] - 1) +
//                        evenPathCount[a[i]] * (odd[index] - even[a[i]] - evenPathCount[a[i]] * value[index] - value[index]) +
//                        2 * (even[a[i]] + value[index] * evenPathCount[a[i]])) % 1000000007
//                ;
                temp = (temp + odd[a[i]] * (evenPathCount[index] - oddPathCount[a[i]])) % 1000000007;
                temp = (temp + oddPathCount[a[i]] * (even[index] - odd[a[i]] + oddPathCount[a[i]] * value[index])) % 1000000007;
                amount = (amount + temp) % 1000000007;
            }
            i = b[i];
        }
        amount = (amount + value[index]) % 1000000007;
        return amount;
    }

}
