Problem:
Score : 300 points

Problem Statement

N persons are standing in a row. The height of the i-th person from the front is A_i.

We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person:

Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool.

Find the minimum total height of the stools needed to meet this goal.

Constraints

1 \leq N \leq 2\times 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 \ldots A_N

Output

Print the minimum total height of the stools needed to meet the goal.

Sample Input 1

5
2 1 5 4 3

Sample Output 1

4

If the persons stand on stools of heights 0, 1, 0, 1, and 2, respectively, their heights will be 2, 2, 5, 5, and 5, satisfying the condition.

We cannot meet the goal with a smaller total height of the stools.

Sample Input 2

5
3 3 3 3 3

Sample Output 2

0

Giving a stool of height 0 to everyone will work.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

//AUTHOR: RAVAN_2070
//PUNE INSTITUTE OF COMPUTER TECHNOLOGY
/*                                                    
      I ♥ CLARICE STARLING...                                                               
        EXPLAINATION BELOW->
*/
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef long double ld;
typedef vector<ll> vi;
typedef vector<pair<ll,ll>> vii;
typedef pair<ll,ll> pii;
typedef map<ll,ll> mii;
#define MOD7 1000000007
#define MOD9 1000000009
#define pi 3.1415926535
#define Test_cases ll TC;cin>>TC;while(TC--)
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(), x.rend()
#define sp(x) fixed<<setprecision(x)
#define sz(x) (ll)(x.size())
#define fo(i,a,b) for(i=a;i<b;i++)
#define foe(i,a,b) for(i=a;i<=b;i++)
const ll INF = (ll)2e18 + 77;
void solve()
{
   int n,i;cin>>n;
   ll sum=0ll;
   ll A[n];
   fo(i,0,n)cin>>A[i];
   ll mx=A[0];
   fo(i,1,n)
   {
       if(mx>A[i])sum+=mx-A[i];
       mx=max(mx,A[i]);
   }
   cout<<sum<<"\n";
}
int main()
{
   fastio
   //Test_cases
   solve();
  return 0;
}
//**always check if you are not killing input in between
//map<ll,ll,greater<int>> to sort in descending order ,in iterator declaration no greater<int>
//if you want to apply lowerbound on set use s.lower_bound(key),not the one used for vector
// Syntax to create a min heap for priority queue priority_queue <int, vector<int>, greater<int>>; 
//for modinverse use modularexponention(x,M-2,M);
//for "each" subtraction from ans(ans-k) if(ans<0) do ans+=MOD,ans%=MOD
//vector<vector<int>>dp(201,vector<int>(n+1)); A way to initialize vector of vector
//string2=string1.substr(pos,length);if len no specified then till end
//void fill (ForwardIterator first, ForwardIterator last, const T& val);
//The range filled is [first,last), which contains all the elements between first and last, including the element pointed by first but not the element pointed by last.


//If wrong on pretest 2 means logically wrong
/* TIPS FOR DEBUGGING
 * Check if values are not overflowing (use long long where required)
 * Check if ranges are inclusive or not.
 * Check properly if greedy will work or not before using it.
 * Check EDGE CASES!! Seriously, even in 2B, saving 8 points by being
 * 2 minutes faster will bite you in the ass if you don't make sure
 * it works on edge cases.
 * Keep adding to this list.
*/