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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class DivCmb_474E {

	static final long MOD = 1_000_000_007;

	static long[] vals;

	static int[] ePt;
	static int[] nE;
	static int[] lSt;

	static int[] mark;
	static int[][] cnt;

	static int[] tCnt;

	static long sum = 0;

	public static void main(String[] args) throws IOException {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter printer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int nV = Integer.parseInt(reader.readLine());

		StringTokenizer inputData = new StringTokenizer(reader.readLine());
		vals = new long[nV];
		for (int i = 0; i < nV; i++) {
			vals[i] = Integer.parseInt(inputData.nextToken());
		}

		ePt = new int[2 * nV];
		nE = new int[2 * nV];
		lSt = new int[nV];
		Arrays.fill(lSt, -1);

		int eI = 0;
		for (int i = 0; i < nV - 1; i++) {
			inputData = new StringTokenizer(reader.readLine());
			int a = Integer.parseInt(inputData.nextToken()) - 1;
			int b = Integer.parseInt(inputData.nextToken()) - 1;
			ePt[eI] = b;
			nE[eI] = lSt[a];
			lSt[a] = eI++;

			ePt[eI] = a;
			nE[eI] = lSt[b];
			lSt[b] = eI++;
		}

		mark = new int[nV];
		cnt = new int[2][nV];

		dfs(0, -1);

		tCnt = new int[] { cnt[0][0], cnt[1][0] };

		calculate(0, -1);
		sum = (sum + MOD) % MOD;
		printer.println(sum);
		printer.close();
	}

	static void dfs(int cV, int pV) {
		cnt[mark[cV]][cV]++;
		int aI = lSt[cV];
		while (aI != -1) {
			int aV = ePt[aI];
			if (aV != pV) {
				mark[aV] = mark[cV] ^ 1;
				dfs(aV, cV);
				cnt[0][cV] += cnt[0][aV];
				cnt[1][cV] += cnt[1][aV];
			}
			aI = nE[aI];
		}
	}

	static void calculate(int cV, int pV) {
		int cMark = mark[cV];

		int aI = lSt[cV];

		long[] cCnt = new long[2];
		cCnt[cMark]++;
		sum = (sum + vals[cV]) % MOD;

		while (aI != -1) {
			int aV = ePt[aI];
			if (aV != pV) {
				sum = (sum + 2 * cCnt[cMark] * cnt[cMark][aV] % MOD * vals[cV]) % MOD;
				sum = (sum - 2 * cCnt[cMark ^ 1] * cnt[cMark ^ 1][aV] % MOD * vals[cV]) % MOD;
				cCnt[0] += cnt[0][aV];
				cCnt[1] += cnt[1][aV];
				calculate(aV, cV);
			}
			aI = nE[aI];
		}
		assert (cCnt[0] == cnt[0][cV] && cCnt[1] == cnt[1][cV]);
		sum = (sum + 2 * cCnt[cMark] * (tCnt[cMark] - cCnt[cMark]) % MOD * vals[cV]) % MOD;
		sum = (sum - 2 * cCnt[cMark ^ 1] * (tCnt[cMark ^ 1] - cCnt[cMark ^ 1]) % MOD * vals[cV]) % MOD;
	}
}
