Problem:
Score : 300 points

Problem Statement

There are N+1 towns. The i-th town is being attacked by A_i monsters.

We have N heroes. The i-th hero can defeat monsters attacking the i-th or (i+1)-th town, for a total of at most B_i monsters.

What is the maximum total number of monsters the heroes can cooperate to defeat?

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_{N+1}
B_1 B_2 ... B_N

Output

Print the maximum total number of monsters the heroes can defeat.

Sample Input 1

2
3 5 2
4 5

Sample Output 1

9

If the heroes choose the monsters to defeat as follows, they can defeat nine monsters in total, which is the maximum result.

The first hero defeats two monsters attacking the first town and two monsters attacking the second town.

The second hero defeats three monsters attacking the second town and two monsters attacking the third town.

Sample Input 2

3
5 6 3 8
5 100 8

Sample Output 2

22

Sample Input 3

2
100 1 1
1 100

Sample Output 3

3

Solution in Ruby (2.3.3) (judge verdict: Accepted)

def s()
  gets().chomp
end
def i()
  gets.to_i
end
def li()
  gets.split.map(&:to_i)
end
def ili()
  n = gets.to_i
  as = gets.split.map(&:to_i)
  [n, as]
end
def inli()
  n = gets.to_i
  as = []
  n.times {
    as << gets.split.map(&:to_i)
  }
  [n, as]
end
class Array
  def sum
    total = 0
    self.each do |item|
      total += item
    end
    total
  end
end

def eratosthenes(size)
  prime_list = []
  is_prime = Array.new(size, true)
  is_prime[0] = false
  is_prime[1] = false

  2.upto(size-1){|i|
    if is_prime[i]
      prime_list << i
      (2).upto((size.to_f/i).ceil-1){|j|
        is_prime[j*i] = false
      }
    end
  }
  [prime_list, is_prime]
end

def dijkstra(s, e, cost)
  size = cost.length
  if s==e
    return 0
  end
  dp = {}
  dp[s] = 0
  undefined_node = {}
  size.times{|i|
    if i != s
      undefined_node[i] = cost[s][i]
    end
  }
  while undefined_node.length != 0
    min = undefined_node.min_by{|_, v| v}
    stock = {}
    temp = undefined_node.clone
    temp.each{|k, v|
      if min[1] == v
        undefined_node.delete(k)
        if k == e
          return v
        end
        dp[k] = v
        stock[k] = v
      end
    }

    stock.each{|k, v|
      undefined_node.each{|k2, v2|
        undefined_node[k2] = v + cost[k][k2] < v2 ?  v + cost[k][k2] : v2
      }
    }
  end
end

def get_bits(len)
  res =  []
  (2**len).times{|i|
    res << i.to_s(2).rjust(len, '0')
  }
  res
end

def get_GCD(m, n)
  tmp = [n, m].max
  n = [n, m].min
  m = tmp
  while n != 0
    tmp = m % n
    m = n
    n = tmp
  end
  m
end


n = i()
as =li()
bs = li()
sum = 0
n.times {|i|
  if as[i] >= bs[i]
    sum += bs[i]
  elsif bs[i]-as[i] > as[i+1]
    sum += as[i] + as[i+1]
    as[i+1] = 0
  else
    sum += bs[i]
    as[i+1] -= (bs[i]-as[i])
  end
}
p sum