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 \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). 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. Print the total sum of alternating functions of all unique simple paths modulo (10^{9}+7). 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), $