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.Scanner;

public class E {
	static long freq[];
	static ArrayList<Integer>[] adj;
	static int [] subtreesz;
	static int [] numEven;
	static int n;
	static int MOD = 1000000007;
	static int [] depth;
	static int even;
	public static void main(String[] args) {
		Scanner sc=  new Scanner(System.in);
		n= sc.nextInt();
		int V[]  = new int[n];
		freq = new long[n];
		for(int i=0;i<n;i++) {
			V[i] = sc.nextInt();
		}
		adj = new ArrayList[n];
		for(int i=0;i<n;i++) {
			adj[i] = new ArrayList<>();
		}
		for(int i=0;i<n-1;i++) {
			int u = sc.nextInt()-1;
			int v = sc.nextInt()-1;
			adj[u].add(v);
			adj[v].add(u);
		}
		subtreesz =new int[n];
		numEven = new int[n];
		depth = new int[n];
		dfsDepth(0,0);
		even = 0;
		for(int i=0;i<n;i++) {
			if(depth[i]%2==1) {
				even++;
			}
		}
		dfs(0,0);
		long ans =0;
		for(int i=0;i<n;i++) {
			freq[i]%=MOD;
			ans+=(long) freq[i]*V[i];
			ans%=MOD;
		}
		if(ans<0) ans+=MOD;
		
		System.out.println(ans);
	}
	static void dfsDepth(int curr,int par) {
		depth[curr] = depth[par]+1;
		for(int next : adj[curr]) {
			if(next!=par) {
				dfsDepth(next,curr);
			}
		}
	}
	static void dfs(int curr,int par) {
		subtreesz[curr]++;
		numEven[curr]++;
		for(int next : adj[curr]) {
			if(next!=par) {
				dfs(next,curr);
				numEven[curr]+=(subtreesz[next]-numEven[next]);
				subtreesz[curr]+=subtreesz[next];
			}
		}
		//inside subtree
		for(int next : adj[curr]) {
			if(next!=par) {
				freq[curr]+=(long) (subtreesz[next]-numEven[next])*(n-subtreesz[next]);
				freq[curr]-=(long) (numEven[next])*(n-subtreesz[next]);
			}
		}
		freq[curr]+=(long) (n); 
		//outside subtree
		if(depth[curr]%2==1) {
			//even from 0
			freq[curr]+=(long) (even-numEven[curr])*subtreesz[curr];
			freq[curr]-=(long) (n-even-(subtreesz[curr]-numEven[curr]))*subtreesz[curr];
		}
		else {
			freq[curr]+=(long) (n-even-numEven[curr])*subtreesz[curr];
			freq[curr]-=(long) (even-(subtreesz[curr]-numEven[curr]))*subtreesz[curr];
		}
	}
}
