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1777D · Score of a Tree

1900 · bitmasks, combinatorics, dfs and similar

Problem: You are given a tree of nn nodes, rooted at 11. Every node has a value of either 00 or 11 at time t=0t=0.

At any integer time t>0t>0, the value of a node becomes the bitwise XOR of the values of its children at time t1t - 1; the values of leaves become 00 since they don't have any children.

Let S(t)S(t) denote the sum of values of all nodes at time tt.

Let F(A)F(A) denote the sum of S(t)S(t) across all values of tt such that 0t101000 \le t \le 10^{100}, where AA is the initial assignment of 00s and 11s in the tree.

The task is to find the sum of F(A)F(A) for all 2n2^n initial configurations of 00s and 11s in the tree. Print the sum modulo 109+710^9+7.

Input Format: Each test contains multiple test cases. The first line contains the number of test cases tt (1t1051 \le t \le 10^5). The description of the test cases follows.

The first line of each test case contains nn (1n21051 \le n \le 2 \cdot 10^5) — the number of nodes in the tree.

The next n1n-1 lines of each test case contain two integers each — uu, vv indicating an edge between uu and vv (1u,vn1 \le u, v \le n).

It is guaranteed that the sum of nn over all test cases does not exceed 21052 \cdot 10^5.

Output Format: Output the sum modulo 109+710^9+7 for each test case.

Note: Let us find F(A)F(A) for the configuration A=[0,1,0,0,1,1]A = [0,1,0,0,1,1] (A[i]A[i] denotes the value of node ii). Initially (at t=0t = 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)S(0) for this configuration is 33.

At t=1t = 1 the configuration changes to [1,0,0,0,0,0][1,0,0,0,0,0]. The tree looks as shown below. S(1)=1S(1) = 1.

At t=2t = 2 the configuration changes to [0,0,0,0,0,0][0,0,0,0,0,0]. The tree looks as shown below. S(2)=0S(2) = 0.

For all t>2t>2, the graph remains unchanged, so S(t)=0S(t)=0 for all t>2t > 2. So, for the initial configuration A=[0,1,0,0,1,1]A = [0,1,0,0,1,1], the value of F(A)=3+1=4F(A) = 3 + 1 = 4.

Doing this process for all possible 262^{6} configurations yields us an answer of 288\textbf{288}.

Sample Cases

Case 1

Input

1
6
1 2
1 3
3 4
3 5
3 6

Output

288

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