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combinatorics

dfs and similar

divide and conquer

dp

probabilities

trees

*2300

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E. Alternating Tree

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputGiven 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 $$$u_1 \rightarrow u_2 \rightarrow u_3 \rightarrow \dots u_{m-1} \rightarrow u_{m}$$$, then its alternating function $$$A(u_{1},u_{m})$$$ is defined as $$$A(u_{1},u_{m}) = \sum\limits_{i=1}^{m} (-1)^{i+1} \cdot V_{u_{i}}$$$. 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 $$$n$$$ $$$(2 \leq n \leq 2\cdot10^{5} )$$$ — the number of vertices in the tree.

The second line contains $$$n$$$ space-separated integers $$$V_1, V_2, \ldots, V_n$$$ ($$$-10^9\leq V_i \leq 10^9$$$) — values of the nodes.

The next $$$n-1$$$ lines each contain two space-separated integers $$$u$$$ and $$$v$$$ $$$(1\leq u, v\leq n, u \neq v)$$$ 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 $$$2$$$: $$$1 \rightarrow 2$$$ has alternating function equal to $$$A(1,2) = 1 \cdot (-4)+(-1) \cdot 1 = -5$$$.

A simple path from node $$$1$$$ to node $$$3$$$: $$$1 \rightarrow 3$$$ has alternating function equal to $$$A(1,3) = 1 \cdot (-4)+(-1) \cdot 5 = -9$$$.

A simple path from node $$$2$$$ to node $$$4$$$: $$$2 \rightarrow 1 \rightarrow 4$$$ has alternating function $$$A(2,4) = 1 \cdot (1)+(-1) \cdot (-4)+1 \cdot (-2) = 3$$$.

A simple path from node $$$1$$$ to node $$$1$$$ has a single node $$$1$$$, so $$$A(1,1) = 1 \cdot (-4) = -4$$$.

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.

Codeforces (c) Copyright 2010-2024 Mike Mirzayanov

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