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)

//g++ t.cpp -o t && t < p.txt
//d>p.txt&&t<p.txt&&t2<p.txt

#include <iostream> // プログラムの耳と口
#include <vector> // 超有能配列秘書
#include <algorithm> // みんな大好きソートと二分探索
#include <queue> // きゅーちゃん、だいすき
#include <string> // to_string
#include <functional> // 関数を変数に入れる子
#include <set> // 値の取得・挿入・削除を高速に
#include <map> // setの妹 これまた優秀
#include <random> // 乱択さん
#include <ctime>
#include <bitset>

#define DB cerr<<"D"<<endl
using namespace std; using ll=long long; using ld=long double; const int INF=1e9; const ll LINF=1e18; const double dINF = 1e18; const ld ldINF = 1e18; const double EPS = 1e-6;
template<typename T, typename U, typename O> void caut(T a, U b, O c){cout<<"("<<a<<","<<b<<","<<c<<") ";} template<typename T, typename U> void caut(T a, U b){cout<<"("<<a<<","<<b<<") ";} template<typename T> void caut(T a){cout<<"("<<a<<") ";};
using P=pair<ll,ll>;
const ll M = 1e9+7;
ll mod_pow(ll x, ll a) { ll an = 1; while(a > 0) { if (a&1) an = an * x % M; x = x * x % M; a >>= 1;} return an;}
void add(ll& x, ll y) {x+=y; x%=M;}; void mul(ll& x, ll y) {x*=y; x%=M;}; template<typename T, typename U> void chmax(T& x, U y) {if (x<y) x=y;}; template<typename T, typename U> void chmin(T& x, U y) {if (x>y) x=y;}
bool vaild(int x, int y, int hh, int ww){return 0<=x&&x<hh&&0<=y&&y<ww;}
int keta(ll a) {int res=0; while(a>0) res+=a%10, a/=10; return res;}
const int up[]={1,-1,0,0}, lf[]={0,0,1,-1};

int main() {
  int n; cin>>n;
  ll a[n]; for(int i=0;i<n;i++) cin>>a[i];
  ll an = 0, mx = 0;
  for(int i=0;i<n;i++) {
    mx = max(a[i], mx);
    an+=mx-a[i];
  }
  cout<<an<<endl;
}
// ・配列の大きさok？　・priority_queueはgreater？　・debug消した？
// ・落ち着いて。提出まで4分待ってね……WJ……1/10……2/10……

// Thank you for making problems and running the contest