output_description: Print the number of the possible colors of the last remaining creature after
  the N creatures repeatedly merge and ultimately become one creature.
  
  * * *
input_description: The input is given from Standard Input in the following format:
  
      
      
      N
      A_1 A_2 … A_N
attempt: import sys
  import heapq
  import re
  from itertools import permutations
  from bisect import bisect_left, bisect_right
  from collections import Counter, deque
  from fractions import gcd
  from math import factorial, sqrt, ceil
  from functools import lru_cache, reduce
  
  INF = 1 << 60
  MOD = 1000000007
  sys.setrecursionlimit(10**7)
  
  
  # UnionFind
  class UnionFind:
      def __init__(self, n):
          self.n = n
          self.parents = [-1] * n
  
      def find(self, x):
          if self.parents[x] < 0:
              return x
          else:
              self.parents[x] = self.find(self.parents[x])
              return self.parents[x]
  
      def union(self, x, y):
          x = self.find(x)
          y = self.find(y)
  
          if x == y:
              return
  
          if self.parents[x] > self.parents[y]:
              x, y = y, x
  
          self.parents[x] += self.parents[y]
          self.parents[y] = x
  
      def size(self, x):
          return -self.parents[self.find(x)]
  
      def same(self, x, y):
          return self.find(x) == self.find(y)
  
      def members(self, x):
          root = self.find(x)
          return [i for i in range(self.n) if self.find(i) == root]
  
      def roots(self):
          return [i for i, x in enumerate(self.parents) if x < 0]
  
      def group_count(self):
          return len(self.roots())
  
      def all_group_members(self):
          return {r: self.members(r) for r in self.roots()}
  
      def __str__(self):
          return "\n".join("{}: {}".format(r, self.members(r)) for r in self.roots())
  
  
  def is_prime(n):
      if n == 1:
          return False
      for i in range(2, int(n**0.5) + 1):
          if n % i == 0:
              return False
      return True
  
  
  # ワーシャルフロイド (任意の2頂点の対に対して最短経路を求める)
  # 計算量n^3 (nは頂点の数)
  def warshall_floyd(d, n):
      # d[i][j]: iからjへの最短距離
      for k in range(n):
          for i in range(n):
              for j in range(n):
                  d[i][j] = min(d[i][j], d[i][k] + d[k][j])
      return d
  
  
  # ダイクストラ
  def dijkstra_heap(s, edge, n):
      # 始点sから各頂点への最短距離
      d = [10**20] * n
      used = [True] * n  # True:未確定
      d[s] = 0
      used[s] = False
      edgelist = []
      for a, b in edge[s]:
          heapq.heappush(edgelist, a * (10**6) + b)
      while len(edgelist):
          minedge = heapq.heappop(edgelist)
          # まだ使われてない頂点の中から最小の距離のものを探す
          if not used[minedge % (10**6)]:
              continue
          v = minedge % (10**6)
          d[v] = minedge // (10**6)
          used[v] = False
          for e in edge[v]:
              if used[e[1]]:
                  heapq.heappush(edgelist, (e[0] + d[v]) * (10**6) + e[1])
      return d
  
  
  # 素因数分解
  def factorization(n):
      arr = []
      temp = n
      for i in range(2, int(-(-(n**0.5) // 1)) + 1):
          if temp % i == 0:
              cnt = 0
              while temp % i == 0:
                  cnt += 1
                  temp //= i
              arr.append([i, cnt])
  
      if temp != 1:
          arr.append([temp, 1])
  
      if arr == []:
          arr.append([n, 1])
  
      return arr
  
  
  # 2数の最小公倍数
  def lcm(x, y):
      return (x * y) // gcd(x, y)
  
  
  # リストの要素の最小公倍数
  def lcm_list(numbers):
      return reduce(lcm, numbers, 1)
  
  
  # リストの要素の最大公約数
  def gcd_list(numbers):
      return reduce(gcd, numbers)
  
  
  # 素数判定
  
  
  # limit以下の素数を列挙
  def eratosthenes(limit):
      A = [i for i in range(2, limit + 1)]
      P = []
  
      while True:
          prime = min(A)
  
          if prime > sqrt(limit):
              break
  
          P.append(prime)
  
          i = 0
          while i < len(A):
              if A[i] % prime == 0:
                  A.pop(i)
                  continue
              i += 1
  
      for a in A:
          P.append(a)
  
      return P
  
  
  # 同じものを含む順列
  def permutation_with_duplicates(L):
  
      if L == []:
          return [[]]
  
      else:
          ret = []
  
          # set（集合）型で重複を削除、ソート
          S = sorted(set(L))
  
          for i in S:
  
              data = L[:]
              data.remove(i)
  
              for j in permutation_with_duplicates(data):
                  ret.append([i] + j)
  
          return ret
  
  
  # ここから書き始める
  n = int(input())
  a = sorted(list(map(int, input().split())))
  c = [0 for i in range(n)]
  for i in range(n):
      if i == 0:
          c[i] - a[i]
      else:
          c[i] = c[i - 1] + a[i - 1]
  # print("c =", c)
  ans = n
  for i in range(n):
      if c[i] != 0 and c[i] * 2 < a[i]:
          ans -= i
          break
  print(ans)
  
problem_description: Statement
  
  Snuke found N strange creatures. Each creature has a fixed color and size. The
  color and size of the i-th creature are represented by i and A_i,
  respectively.
  
  Every creature can absorb another creature whose size is at most twice the
  size of itself. When a creature of size A and color B absorbs another creature
  of size C and color D (C \leq 2 \times A), they will merge into one creature
  of size A+C and color B. Here, depending on the sizes of two creatures, it is
  possible that both of them can absorb the other.
  
  Snuke has been watching these creatures merge over and over and ultimately
  become one creature. Find the number of the possible colors of this creature.
samples: [{"input": "3\n    3 1 4", "output": "2\n    \n\nThe possible colors of the last remaining creature are colors 1 and 3. For\nexample, when the creature of color 3 absorbs the creature of color 2, then\nthe creature of color 1 absorbs the creature of color 3, the color of the last\nremaining creature will be color 1.\n\n* * *"}, {"input": "5\n    1 1 1 1 1", "output": "5\n    \n\nThere may be multiple creatures of the same size.\n\n* * *"}, {"input": "6\n    40 1 30 2 7 20", "output": "4"}]