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)

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
#include<bits/stdc++.h>
using namespace std;
using namespace __gnu_pbds;


typedef tree<int, null_type, less<int>, rb_tree_tag,
             tree_order_statistics_node_update>
    ordered_set;


#define pf                  printf
#define sf(a)               scanf("%d",&a)
#define sfl(a)              scanf("%lld",&a)
#define sff(a,b)            scanf("%d %d",&a,&b)
#define sffl(a,b)           scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)         scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)        scanf("%lld %lld %lld",&a,&b,&c)
#define sffff(a,b,c,d)      scanf("%d %d %d %d",&a,&b,&c,&d)
#define sffffl(a,b,c,d)     scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sfffff(a,b,c,d,e)   scanf("%d %d %d %d %d",&a,&b,&c,&d,&e)
#define sfffffl(a,b,c,d,e)  scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)
#define sfc(a)              scanf("%c",&a)
#define pii                 pair<int,int>
#define ms(a,b)             memset(a,b,sizeof(a))
#define pb(a)               push_back(a)
#define pbp(a,b)            push_back({a,b})
#define db                  double
#define ft                  float
#define ll                  long long
#define ull                 unsigned long long
#define pii                 pair<int,int>
#define ff                  first
#define ss                  second
#define sz(x)               x.size()
#define all(x)              x.begin(),x.end()
#define CIN                 ios_base::sync_with_stdio(0); cin.tie(0)
#define max3(a, b, c)       max(a, b) > max(b, c) ? max(a, b) : max(b, c)
#define min3(a, b, c)       min(a, b) < min(b, c) ? min(a, b) : min(b, c)
#define for0(i,n)           for(int i=0;i<n;i++)
#define for1(i,n)           for(int i=1;i<=n;i++)
#define forrev(i,n)         for(int i=n-1; i>=0; i--)
#define forab(i,a,b)        for(int i=a;i<=b;i++)
#define forba(i,b,a)        for(int i=b;i>=a;i--)
#define stlloop(x)          for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define gcd(a, b)           __gcd(a, b)
#define lcm(a, b)           ((a)*((b)/gcd(a,b)))
#define case1(z)            cout<<"Case "<<z<<": "
#define case2(z)            printf("Case %d: ",z)
#define PI                  acos(-1) //3.14159265358979323846264338328
#define valid(tx,ty)        tx>=0 && tx<row && ty>=0 && ty<col
#define intlim              2147483648
#define N                   10000007
#define inf                 100000008
#define mod                 1000000007

/*------------------------------Graph Moves----------------------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*---------------------------------------------------------------------*/

/*
vector<ll>vec;
bool chk[N];
void sieve()
{
    ll i,j;
    for(i=4;i<N;i+=2)
        chk[i]=1;
    for(i=3;i<=sqrt(N);i+=2)
    {
        for(j=i*i;j<N;j+=2*i)
            chk[j]=1;
    }
    for(i=2;i<N;i++)
    {
        if(chk[i]==0)
        {
            vec.pb(i);
        }
    }
}
*/


ll a[N];


int main()
{
    ll n,i,mx=-1,ans=0;
    cin>>n;
    for1(i,n)
    {
        cin>>a[i];
    }
    for1(i,n)
    {
        if(a[i]>mx)
        {
            mx=a[i];
        }
        ans+=(mx-a[i]);
    }
    cout<<ans<<endl;
    return 0;
}