The Little Elephant loves sortings.
He has an array a consisting of n integers. Let's number the array elements from 1 to n, then the i-th element will be denoted as ai. The Little Elephant can make one move to choose an arbitrary pair of integers l and r (1 ≤ l ≤ r ≤ n) and increase ai by 1 for all i such that l ≤ i ≤ r.
Help the Little Elephant find the minimum number of moves he needs to convert array a to an arbitrary array sorted in the non-decreasing order. Array a, consisting of n elements, is sorted in the non-decreasing order if for any i (1 ≤ i < n) ai ≤ ai + 1 holds.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the size of array a. The next line contains n integers, separated by single spaces — array a (1 ≤ ai ≤ 109). The array elements are listed in the line in the order of their index's increasing.
Output
In a single line print a single integer — the answer to the problem.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
3 1 2 3
Output
0
Input
3 3 2 1
Output
2
Input
4 7 4 1 47
Output
6
Note
In the first sample the array is already sorted in the non-decreasing order, so the answer is 0.
In the second sample you need to perform two operations: first increase numbers from second to third (after that the array will be: [3, 3, 2]), and second increase only the last element (the array will be: [3, 3, 3]).
In the third sample you should make at least 6 steps. The possible sequence of the operations is: (2; 3), (2; 3), (2; 3), (3; 3), (3; 3), (3; 3). After that the array converts to [7, 7, 7, 47].
题解：记录之前最高的，若当前比前一个高比之前最高矮，让之前的先都加到当前这个，若比之前最高的还高，让之前都增加到之前最高
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define lowbit(x) (x&(-x))
#define INF 0x3f3f3f3f
const int N=110100;
typedef long long ll;
ll n;
ll a[N];
int main()
{
ll x,l;
while(scanf("%lld",&n)!=EOF)
{
ll ans=0;
scanf("%d",&a[1]);
l=a[1];
for(int i=2;i<=n;i++)
{
scanf("%d",&a[i]);
if(a[i]<=l&&a[i]>a[i-1])
ans+=a[i]-a[i-1];
else if(a[i]>l)
{
ans+=l-a[i-1];
l=a[i];
}
}
if(a[n]<l) ans+=l-a[n];
cout<<ans<<endl;
}
return 0;
}