Problem:
Score : 300 points

Problem Statement

In a flower bed, there are N flowers, numbered 1,2,......,N. Initially, the heights of all flowers are 0.
You are given a sequence h=\{h_1,h_2,h_3,......\} as input. You would like to change the height of Flower k to h_k for all k (1 \leq k \leq N), by repeating the following "watering" operation:

Specify integers l and r. Increase the height of Flower x by 1 for all x such that l \leq x \leq r.

Find the minimum number of watering operations required to satisfy the condition.

Constraints

1 \leq N \leq 100

0 \leq h_i \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
h_1 h_2 h_3 ...... h_N

Output

Print the minimum number of watering operations required to satisfy the condition.

Sample Input 1

4
1 2 2 1

Sample Output 1

2

The minimum number of watering operations required is 2.
One way to achieve it is:

Perform the operation with (l,r)=(1,3).

Perform the operation with (l,r)=(2,4).

Sample Input 2

5
3 1 2 3 1

Sample Output 2

5

Sample Input 3

8
4 23 75 0 23 96 50 100

Sample Output 3

221

Solution in Lua (5.3.2) (judge verdict: Runtime Error)

local N=io.read("n")
h={}
for i=1,N do 
  h[i]=io.read("n")
  end
a=1
s=0
while a<=N-s do
  if h[a]==h[a+1] then
    for i=a,N-1-s do 
      h[i]=h[i+1]
      end
    s=s+1
    else
    a=a+1
    end
  end
if N-s==1 then 
  print(h[1])
  end
if N-s==2 then 
  if h[1]>h[2] then
    print(h[1])
    else print(h[2])
    end
end
D=0
d=0
if h[1]>h[2] then
  D=D+h[1]
  end
if h[N-s]>h[N-s-1] then
  D=D+h[N-s]
  end
for i=2,N-s-1 do 
  if h[i]>h[i-1] and h[i]>h[i+1] then
  D=D+h[i]
  elseif h[i]<h[i-1] and h[i]<h[i+1] then
  d=d+h[i]
  end
end
print(D-d)