阅读以下用动态规划解决的0-1背包问题的函数，假设背包的容量 是10kg，假设输入4个物品的重量分别为 （单位为kg），每个物品对应的价值 分别为 ，则函数的输出为（    ）。

def knapsack(capacity, weights, values):

      dp = [[0 for _ in range(capacity + 1)] for _ in range(len(weights) + 1)]

      for i in range(1, len(weights) + 1):

            for j in range(1, capacity + 1):

                  if weights[i - 1] <= j:

                        dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weights[i - 1]] + values[i - 1])

                  else:

                        dp[i][j] = dp[i - 1][j]

      return dp[-1][-1]

weights = [1, 3, 4,6]

values = [20,30,50,60]

capacity = 10

print(knapsack(capacity, weights, values))