Buggy python submission (verdict: Time Limit Exceeded):
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline

N = int(input())
MOD = 10**9 + 7
tree = [[] for i in range(N+1)]
for i in range(N-1):
  a, b = map(int, input().split())
  tree[a].append(b)
  tree[b].append(a)

# 頂点1をルートにして各頂点より下にある部分木の頂点数を求めていく
root, par = 1, 0
sizes = [0] * (N+1)
order = [[root, par]]
def calc_nums(node, par):
  global sizes
  global order
  if sizes[node] > 0:
    return sizes[node]
  children = tree[node]
  res = 1
  for child in children:
    if child == par:
      continue
    order.append([child, node])
    res += calc_nums(child, node)
  sizes[node] = res
  return res

_ = calc_nums(root, par)


# 各頂点について、その頂点が穴あき度としてカウントされる状況を独立に計算する
demo = 0 # 分子
div = pow(2, N, MOD) # 分母
base = pow(2, N-1, MOD) - 1
for node, par in order:
  children = tree[node]
  num = base
  num_par = N - 1
  for child in children:
    if child == par:
      continue
    s = sizes[child]
    num_par -= s
    num -= pow(2, s, MOD) - 1
  num -= pow(2, num_par, MOD) - 1
  demo += num
    
ans = demo * pow(div, MOD-2, MOD) % MOD
print(ans)

Fixed submission:
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline

N = int(input())
MOD = 10**9 + 7
tree = [[] for i in range(N+1)]
for i in range(N-1):
  a, b = map(int, input().split())
  tree[a].append(b)
  tree[b].append(a)

# 頂点1をルートにして各頂点より下にある部分木の頂点数を求めていく
root, par = 1, 0
sizes = [0] * (N+1)
order = [[root, par]]
def calc_nums(node, par):
  global sizes
  global order
  if sizes[node] > 0:
    return sizes[node]
  children = tree[node]
  res = 1
  for child in children:
    if child == par:
      continue
    order.append([child, node])
    res += calc_nums(child, node)
  sizes[node] = res
  return res

_ = calc_nums(root, par)

# 2の累乗を事前に計算しておく
prod2 = [1]
for i in range(N):
  prod2.append(prod2[-1] * 2 % MOD)

# 各頂点について、その頂点が穴あき度としてカウントされる状況を独立に計算する
demo = 0 # 分子
div = prod2[N] # 分母
base = prod2[N-1] - 1
for node, par in order:
  children = tree[node]
  num = base
  num_par = N - 1
  for child in children:
    if child == par:
      continue
    s = sizes[child]
    num_par -= s
    num -= prod2[s] - 1 
  num -= prod2[num_par] - 1
  demo += num
    
ans = demo * pow(div, MOD-2, MOD) % MOD
print(ans)

Buggy python submission (verdict: Time Limit Exceeded):
import sys
sys.setrecursionlimit(10**9)

N = int(input())
MOD = 10**9 + 7
tree = [[] for i in range(N+1)]
for i in range(N-1):
  a, b = map(int, input().split())
  tree[a].append(b)
  tree[b].append(a)

# 頂点1をルートにして各頂点より下にある部分木の頂点数を求めていく
root, par = 1, 0
sizes = [0] * (N+1)
order = [[root, par]]
def calc_nums(node, par):
  global sizes
  global order
  if sizes[node] > 0:
    return sizes[node]
  children = tree[node]
  res = 1
  for child in children:
    if child == par:
      continue
    order.append([child, node])
    res += calc_nums(child, node)
  sizes[node] = res
  return res

calc_nums(root, par)

# 2の累乗を事前に計算しておく
prod2 = [1]
for i in range(N):
  prod2.append(prod2[-1] * 2 % MOD)

# 各頂点について、その頂点が穴あき度としてカウントされる状況を独立に計算する
demo = 0 # 分子
div = prod2[N] # 分母
base = prod2[N-1] - 1
for node, par in order:
  children = tree[node]
  num = base
  num_par = N - 1
  for child in children:
    if child == par:
      continue
    s = sizes[child]
    num_par -= s
    num -= prod2[s] - 1 
  num -= prod2[num_par] - 1
  demo += num
    
ans = demo * pow(div, MOD-2, MOD) % MOD
print(ans)

Fixed submission:
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline

N = int(input())
MOD = 10**9 + 7
tree = [[] for i in range(N+1)]
for i in range(N-1):
  a, b = map(int, input().split())
  tree[a].append(b)
  tree[b].append(a)

# 頂点1をルートにして各頂点より下にある部分木の頂点数を求めていく
root, par = 1, 0
sizes = [0] * (N+1)
order = [[root, par]]
def calc_nums(node, par):
  global sizes
  global order
  if sizes[node] > 0:
    return sizes[node]
  children = tree[node]
  res = 1
  for child in children:
    if child == par:
      continue
    order.append([child, node])
    res += calc_nums(child, node)
  sizes[node] = res
  return res

_ = calc_nums(root, par)

# 2の累乗を事前に計算しておく
prod2 = [1]
for i in range(N):
  prod2.append(prod2[-1] * 2 % MOD)

# 各頂点について、その頂点が穴あき度としてカウントされる状況を独立に計算する
demo = 0 # 分子
div = prod2[N] # 分母
base = prod2[N-1] - 1
for node, par in order:
  children = tree[node]
  num = base
  num_par = N - 1
  for child in children:
    if child == par:
      continue
    s = sizes[child]
    num_par -= s
    num -= prod2[s] - 1 
  num -= prod2[num_par] - 1
  demo += num
    
ans = demo * pow(div, MOD-2, MOD) % MOD
print(ans)

Buggy python submission (verdict: Time Limit Exceeded):
import sys
sys.setrecursionlimit(10**9)

N = int(input())
MOD = 10**9 + 7
tree = [[] for i in range(N+1)]
for i in range(N-1):
  a, b = map(int, input().split())
  tree[a].append(b)
  tree[b].append(a)

# 頂点1をルートにして各頂点より下にある部分木の頂点数を求めていく
root, par = 1, 0
sizes = [0] * (N+1)
order = [[root, par]]
def calc_nums(node, par):
  global sizes
  global order
  if sizes[node] > 0:
    return sizes[node]
  children = tree[node]
  res = 1
  for child in children:
    if child == par:
      continue
    order.append([child, node])
    res += calc_nums(child, node)
  sizes[node] = res
  return res

calc_nums(root, par)

# 2の累乗を事前に計算しておく
prod2 = [1]
for i in range(N):
  prod2.append(prod2[-1] * 2 % MOD)

# 各頂点について、その頂点が穴あき度としてカウントされる状況を独立に計算する
demo = 0 # 分子
div = prod2[N] # 分母
base = prod2[N-1] - 1
for node, par in order:
  children = tree[node]
  num = base
  num_par = N - 1
  for child in children:
    if child == par:
      continue
    s = sizes[child]
    num_par -= s
    num -= prod2[s] - 1 
  num -= prod2[num_par] - 1
  demo += num
    
ans = demo * pow(div, MOD-2, MOD) % MOD
print(ans)

Fixed submission:
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline

N = int(input())
MOD = 10**9 + 7
tree = [[] for i in range(N+1)]
for i in range(N-1):
  a, b = map(int, input().split())
  tree[a].append(b)
  tree[b].append(a)

# 頂点1をルートにして各頂点より下にある部分木の頂点数を求めていく
root, par = 1, 0
sizes = [0] * (N+1)
order = [[root, par]]
def calc_nums(node, par):
  global sizes
  global order
  if sizes[node] > 0:
    return sizes[node]
  children = tree[node]
  res = 1
  for child in children:
    if child == par:
      continue
    order.append([child, node])
    res += calc_nums(child, node)
  sizes[node] = res
  return res

_ = calc_nums(root, par)

# 2の累乗を事前に計算しておく
prod2 = [1]
for i in range(N):
  prod2.append(prod2[-1] * 2 % MOD)

# 各頂点について、その頂点が穴あき度としてカウントされる状況を独立に計算する
demo = 0 # 分子
div = prod2[N] # 分母
base = prod2[N-1] - 1
for node, par in order:
  children = tree[node]
  num = base
  num_par = N - 1
  for child in children:
    if child == par:
      continue
    s = sizes[child]
    num_par -= s
    num -= prod2[s] - 1 
  num -= prod2[num_par] - 1
  demo += num
    
ans = demo * pow(div, MOD-2, MOD) % MOD
print(ans)

Buggy python submission (verdict: Runtime Error):
N = int(input())
MOD = 10**9 + 7
tree = [[] for i in range(N+1)]
for i in range(N-1):
  a, b = map(int, input().split())
  tree[a].append(b)
  tree[b].append(a)

# 頂点1をルートにして各頂点より下にある部分木の頂点数を求めていく
root, par = 1, 0
sizes = [0] * (N+1)
order = [[root, par]]
def calc_nums(node, par):
  global sizes
  global order
  if sizes[node] > 0:
    return sizes[node]
  children = tree[node]
  res = 1
  for child in children:
    if child == par:
      continue
    order.append([child, node])
    res += calc_nums(child, node)
  sizes[node] = res
  return res

calc_nums(root, par)

# 2の累乗を事前に計算しておく
prod2 = [1]
for i in range(N):
  prod2.append(prod2[-1] * 2 % MOD)

# 各頂点について、その頂点が穴あき度としてカウントされる状況を独立に計算する
demo = 0 # 分子
div = prod2[N] # 分母
base = prod2[N-1] - 1
for node, par in order:
  children = tree[node]
  num = base
  num_par = N - 1
  for child in children:
    if child == par:
      continue
    s = sizes[child]
    num_par -= s
    num -= prod2[s] - 1 
  num -= prod2[num_par] - 1
  demo += num
    
ans = demo * pow(div, MOD-2, MOD) % MOD
print(ans)

Fixed submission:
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline

N = int(input())
MOD = 10**9 + 7
tree = [[] for i in range(N+1)]
for i in range(N-1):
  a, b = map(int, input().split())
  tree[a].append(b)
  tree[b].append(a)

# 頂点1をルートにして各頂点より下にある部分木の頂点数を求めていく
root, par = 1, 0
sizes = [0] * (N+1)
order = [[root, par]]
def calc_nums(node, par):
  global sizes
  global order
  if sizes[node] > 0:
    return sizes[node]
  children = tree[node]
  res = 1
  for child in children:
    if child == par:
      continue
    order.append([child, node])
    res += calc_nums(child, node)
  sizes[node] = res
  return res

_ = calc_nums(root, par)

# 2の累乗を事前に計算しておく
prod2 = [1]
for i in range(N):
  prod2.append(prod2[-1] * 2 % MOD)

# 各頂点について、その頂点が穴あき度としてカウントされる状況を独立に計算する
demo = 0 # 分子
div = prod2[N] # 分母
base = prod2[N-1] - 1
for node, par in order:
  children = tree[node]
  num = base
  num_par = N - 1
  for child in children:
    if child == par:
      continue
    s = sizes[child]
    num_par -= s
    num -= prod2[s] - 1 
  num -= prod2[num_par] - 1
  demo += num
    
ans = demo * pow(div, MOD-2, MOD) % MOD
print(ans)