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())
peer = [[] for _ in range(n)]
for _ in range(n-1):
  a,b = map(int,input().split())
  a -= 1
  b -= 1
  peer[a].append(b)
  peer[b].append(a)
pre = [10**6 for _ in range(n)]
pre[0] = -1
pst = [[] for _ in range(n)]
rank = [[0]]
seen = [0 for _ in range(n)]
seen[0] = 1
now = [0]
while now:
  last = now
  now = []
  for x in last:
    for y in peer[x]:
      if seen[y] == 0:
        seen[y] = 1
        now.append(y)
        pre[y] = x
        pst[x].append(y)
  rank.append(now)
rank = rank[1:-1]
#print(rank)

family = [1 for _ in range(n)]
rank.reverse()
for z in rank:
  for w in z:
    family[pre[w]] += family[w]
#print(pre)
#print(family)

mod = 10**9+7
half = 5*10**8+4
satoshi = 1
btc = [1]
for _ in range(n):
  satoshi *= half
  satoshi %= mod
  btc.append(btc[-1]*2%mod)
#print(satoshi,btc)

ans = 0
for i in range(n):
  if i == 0:
    if len(pst[i]) == 1:
      pass
    else:
      q = []
      for x in pst[i]:
        q.append(family[x])
      tmp = -len(q)+1
      for p in q:
        tmp += btc[p]
      tmp = btc[n-1] - tmp
      #print(tmp)
      ans += tmp
  else:
    if len(pst[i]) == 0:
      pass
    else:
      q = []
      for x in pst[i]:
        q.append(family[x])
      q.append(n-1-sum(q))
      tmp = -len(q)+1
      for p in q:
        tmp += btc[p]
      tmp = btc[n-1] - tmp
      #print(tmp)
      ans += tmp
  #try:
    #print(i,q)
  #except:
    #pass
ans *= satoshi
ans %= mod
print(ans)