B"You are given a tree of n nodes, rooted at 1 . Every node has a value of either 0 or 1 at time t=0 . At any integer time t>0 , the value of a node becomes the bitwise XOR of the values of its children at time t - 1 ; the values of leaves become 0 since they don't have any children. Let S(t) denote the sum of values of all nodes at time t . Let F(A) denote the sum of S(t) across all values of t such that 0 <= t <= 10^{100} , where A is the initial assignment of 0 s and 1 s in the tree. The task is to find the sum of F(A) for all 2^n initial configurations of 0 s and 1 s in the tree. Print the sum modulo 10^9+7 . Each test contains multiple test cases. The first line contains the number of test cases t ( 1 <= t <= 10^5 ). The description of the test cases follows. The first line of each test case contains n ( 1 <= n <= 2 cdot 10^5 ) -- the number of nodes in the tree. The next n-1 lines of each test case contain two integers each -- u , v indicating an edge between u and v ( 1 <= u, v <= n ). It is guaranteed that the sum of n over all test cases does not exceed 2 cdot 10^5 . Output the sum modulo 10^9+7 for each test case. Let us find F(A) for the configuration A = [0,1,0,0,1,1] ( A[i] denotes the value of node i ). Initially (at t = 0 ) our tree is as shown in the picture below. In each node, two values are shown: the number and the value of this node. S(0) for this configuration is 3 . At t = 1 the configuration changes to [1,0,0,0,0,0] . The tree looks as shown below. S(1) = 1 . At t = 2 the configuration changes to [0,0,0,0,0,0] . The tree looks as shown below. S(2) = 0 . For all t>2 , the graph remains unchanged, so S(t)=0 for all t > 2 . So, for the initial config"...