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.*;
import java.io.*;

public class E2018 {
    static int n; static long mod = (long) 1e9 + 7; static ArrayList<Integer> [] adj;
    static long [] v; static long [] sz; static long [] tree; static long [] dp;
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        n = sc.nextInt();
        adj = new ArrayList[n+1];
        for (int i = 1; i <= n; i++) adj[i] = new ArrayList<>();
        v = new long[n+1]; sz = new long[n+1]; tree = new long[n+1]; dp = new long[n+1];
        for (int i = 1; i <= n; i++) v[i] = sc.nextLong();
        for (int i = 0; i < n - 1; i++) {
            int a = sc.nextInt(); int b = sc.nextInt();
            adj[a].add(b); adj[b].add(a);
        }
        dfs1(1, -1);
        dp[1] = tree[1];
        dfs2(1, -1);
        long ans = 0;
        for (int i = 1; i <= n; i++) {
            ans = (ans + dp[i]) % mod;
        }
        if (ans < 0) ans += mod;
        out.println(ans);
        out.close();
    }

    static void dfs1(int cur, int par) {
        sz[cur] = 1;
        for (Integer next: adj[cur]) {
            if (next != par) {
                dfs1(next, cur);
                sz[cur] += sz[next];
            }
        }
        tree[cur] = (sz[cur] * v[cur]) % mod;
        for (Integer next: adj[cur]) {
            if (next != par) {
                tree[cur] = (tree[cur] - tree[next]) % mod;
            }
        }
    }

    static void dfs2(int cur, int par) {
        if (par != -1) {
            dp[cur] = (((n - sz[cur]) * v[cur]) % mod);
            dp[cur] = (dp[cur] - dp[par]) % mod;
            if (dp[cur] < 0) dp[cur] += mod;
            dp[cur] = (dp[cur] + (sz[cur] * v[par]) % mod) % mod;
        }
        for (Integer next: adj[cur]) {
            if (next != par) {
                dfs2(next, cur);
            }
        }
    }


    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }


    }

}