Given is a tree T with N vertices. The i-th edge connects Vertex A_i and B_i (1 \leq A_i,B_i \leq N).

Now, each vertex is painted black with probability 1/2 and white with probability 1/2, which is chosen independently from other vertices. Then, let S be the smallest subtree (connected subgraph) of T containing all the vertices painted black. (If no vertex is painted black, S is the empty graph.)

Let the holeyness of S be the number of white vertices contained in S. Find the expected holeyness of S.

Since the answer is a rational number, we ask you to print it \bmod 10^9+7, as described in Notes.

Constraints

* 2 \leq N \leq 2 \times 10^5
* 1 \leq A_i,B_i \leq N
* The given graph is a tree.

Input

Input is given from Standard Input in the following format:


N
A_1 B_1
:
A_{N-1} B_{N-1}


Output

Print the expected holeyness of S, \bmod 10^9+7.

Examples

Input

3
1 2
2 3


Output

125000001


Input

4
1 2
2 3
3 4


Output

375000003


Input

4
1 2
1 3
1 4


Output

250000002


Input

7
4 7
3 1
2 6
5 2
7 1
2 7


Output

570312505
N=int(input())
E=[[] for _ in range(N)]
for _ in range(N-1):
    A,B=map(int,input().split())
    E[A-1].append(B-1)
    E[B-1].append(A-1)
MOD=10**9+7

#dp=子ノードの数+1(自分)
dp=[1]*N

root=0
parent=[0]*N
order=[]
def dfs():
    parent[root]=root
    stack=[root]
    while stack:
        node=stack.pop()
        order.append(node)
        for ss in E[node]:
            if parent[node]!=ss:
                stack.append(ss)
                parent[ss]=node

pow2=[1]*(N+1)
for i in range(1,len(pow2)):
    pow2[i]=pow2[i-1]*2%MOD

def solve():
    ans=0
    inv2=pow(2,MOD-2,MOD)
    #p0=すべての部分木に黒頂点が1つも含まれない確率
    p0=pow(pow2[N-1],MOD-2,MOD)
    for i in reversed(order):
        if root!=i:
            dp[parent[i]]+=dp[i]
        p=0
        if len(E[i])>=2:
            #p0*(p1-1)=どれか1つの各部分木に黒頂点が1つ以上含まれる確率
            p1=1
            for s in E[i]:
                size=0
                if parent[i]!=s:
                    size=dp[s]
                elif i!=root:
                    size=N-dp[i]
                else:
                    continue
                p1+=pow2[size]-1
            p=1-(p0*p1)
            p%=MOD
        ans+=p
        ans%=MOD
    #頂点iが白の確率
    ans=ans*inv2%MOD
    print(ans)

dfs()
solve()
