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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * @author Don Li
 */
public class AlternatingTree {
    
    int MOD = (int) 1e9 + 7;
    int N = (int) 2e5 + 10;
    
    int n;
    int[] val = new int[N];
    int[] sz = new int[N], odd = new int[N], even = new int[N], dep = new int[N], par = new int[N];
    
    int[][] G;
    
    void solve() {
        n = in.nextInt();
        for (int i = 0; i < n; i++) val[i] = in.nextInt();
        int[] fr = new int[n - 1], to = new int[n - 1];
        for (int i = 0; i < n - 1; i++) {
            fr[i] = in.nextInt() - 1;
            to[i] = in.nextInt() - 1;
        }
        G = build_graph(n, fr, to);
        
        dfs(0, -1, 0);
        
        long ans = 0;
        for (int i = 0; i < n; i++) {
            ans = (ans + val[i] * calc(i) % MOD + MOD) % MOD;
        }
        out.println(ans);
    }
    
    long calc(int u) {
        long cnt = n;
        for (int v : G[u]) {
            int even_start, odd_start, end;
            if (v != par[u]) {
                even_start = even[v]; odd_start = odd[v]; end = n - sz[v];
            } else {
                even_start = even[0] - even[u]; odd_start = odd[0] - odd[u]; end = sz[u];
            }
            if (dep[u] % 2 == 0) {
                cnt = add(cnt, (long) (even_start - odd_start) * end);
            } else {
                cnt = add(cnt, (long) (odd_start - even_start) * end);
            }
        }
        return cnt;
    }
    
    long add(long a, long b) {
        return ((a + b) % MOD + MOD) % MOD;
    }
    
    void dfs(int u, int p, int d) {
        par[u] = p;
        dep[u] = d;
        for (int v : G[u]) {
            if (v != p) {
                dfs(v, u, d + 1);
                sz[u] += sz[v];
                odd[u] += odd[v];
                even[u] += even[v];
            }
        }
        sz[u]++;
        if (d % 2 == 0) even[u]++;
        else odd[u]++;
    }
    
    int[][] build_graph(int n, int[] from, int[] to) {
        int[][] G = new int[n][];
        int[] cnt = new int[n];
        for (int i = 0; i < from.length; i++) {
            cnt[from[i]]++;
            cnt[to[i]]++;
        }
        for (int i = 0; i < n; i++) G[i] = new int[cnt[i]];
        for (int i = 0; i < from.length; i++) {
            G[from[i]][--cnt[from[i]]] = to[i];
            G[to[i]][--cnt[to[i]]] = from[i];
        }
        return G;
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new AlternatingTree().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}
