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.*;
import java.util.*;

public class E{
	
	static int n;
	static List<Integer>[] adj;
	static int[] odd;
	static int[] even;
	static int[] bip;
	static long[] v;
	static long mod = (long)1e9+7;
	static long ans;
	static void color(int i, int p, int c) {
		if (bip[i] != -1) return;
		bip[i] = c;
		for (int x: adj[i]) {
			if (x==p) continue;
			color(x,i,(c+1)%2);
		}
	}
	
	static void count(int i, int p) {
		// Count odds and evens per subtree, NOT INCLUDING ITSELF!
		for (int x: adj[i]) {
			if(x==p) continue;
			count(x,i);
			if (bip[x] == 0) even[i]++;
			else odd[i]++;
			odd[i]+=odd[x];
			even[i]+=even[x];
		}
		
	}
	
	static long solveCount(List<Integer> count) {
		long pref = 0;
		long res = 0;
		for (int x: count) pref+=x;
		for (int x: count) {
			pref-=x;
			res+=x*pref;
			res%=mod;
		}
		return res;
	}
	
	static int numEvens, numOdds;
	
	static void calc(int i, int p) {
		List<Integer> zeroCount = new ArrayList<Integer>();
		List<Integer> oneCount = new ArrayList<Integer>();
		zeroCount.add(numEvens-even[i]-(bip[i]==0?1:0));
		oneCount.add(numOdds-odd[i] - (bip[i]==1?1:0));
		for (int x: adj[i]) {
			if(x==p) continue;
			zeroCount.add(even[x] + (bip[x]==0?1:0));
			oneCount.add(odd[x] + (bip[x]==1?1:0));
			calc(x,i);
		}
		long bias = 0; // For the i-i case
		// Calculate all pairwise somehow
		// Handle same bipartite case differently. Include sum of same subtrees as well.
		// Odd paths WILL cancel each other paths. Only even paths matter, and double them.
		// Therefore try to only consider how many evens and odds funnel through this node.
		if (bip[i] == 0) for (int x: zeroCount) bias+=x;
		else for (int x: oneCount) bias+=x;
		//System.out.println("My initial bias: " + bias);
		// Alright, now big brain shit
		//System.out.println(zeroCount + " " + oneCount);
		//System.out.println(solveCount(zeroCount) + " " + solveCount(oneCount));
		
		long zc = solveCount(zeroCount);
		long oc = solveCount(oneCount);
		if (bip[i] == 0) {
			bias+=zc-oc;
		}
		else bias+=oc-zc;
		bias=bias*2+1;
		bias%=mod;
		bias = bias * v[i];
		bias %= mod;
		
		//System.out.println(bias);
		
		ans = (ans + bias + mod)%mod;
	}
	
	public static void main(String[] args) throws IOException {
		// br = new BufferedReader(new FileReader(".in"));
		// out = new PrintWriter(new FileWriter(".out"));
		//new Thread(null, new (), "peepee", 1<<28).start();
		n = readInt();
		adj = new List[n];
		for (int i = 0; i < n; i++) adj[i] = new ArrayList<Integer>();
		odd = new int[n];
		even = new int[n];
		bip = new int[n];
		v = new long[n];
		for (int i = 0; i < n; i++) v[i] =readInt();
		for (int i = 0; i < n-1; i++) {
			int x =readInt()-1;
			int y =readInt()-1;
			adj[x].add(y);
			adj[y].add(x);
		}
		Arrays.fill(bip, -1);
		color(0,0,0);
		count(0,0);
		for (int i = 0; i < bip.length; i++) if (bip[i] == 0) numEvens++; else numOdds++;
		calc(0,0);
		
		out.println(ans);
		out.close();
	}
	
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	static StringTokenizer st = new StringTokenizer("");
	static String read() throws IOException{return st.hasMoreTokens() ? st.nextToken():(st = new StringTokenizer(br.readLine())).nextToken();}
	static int readInt() throws IOException{return Integer.parseInt(read());}
	static long readLong() throws IOException{return Long.parseLong(read());}
	static double readDouble() throws IOException{return Double.parseDouble(read());}
	
}