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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <utility>
#include <tuple>
#include <cstdint>
#include <cstdio>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <deque>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <cctype>
#include <functional>
#include <ctime>
#include <fstream>
#include <cmath>
#include <limits>
#include <numeric>
#include <type_traits>
#include <iomanip>
#include <float.h>
#include <math.h>
#include <cassert>
#pragma warning (disable: 4996)
using namespace std;

using ll = long long;
unsigned euclidean_gcd(unsigned a, unsigned b) {
    if (a < b) return euclidean_gcd(b, a);
    unsigned r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

ll ll_gcd(ll a, ll b) {
    if (a < b) return ll_gcd(b, a);
    ll r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

struct UnionFind {
    vector <ll> par;
    vector <ll> siz;
    UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL) {
        for (ll i = 0; i < sz_; ++i) par[i] = i;
    }
    void init(ll sz_) {
        par.resize(sz_);
        siz.assign(sz_, 1LL);
        for (ll i = 0; i < sz_; ++i) par[i] = i;
    }
    ll root(ll x) {
        while (par[x] != x) {
            x = par[x] = par[par[x]];
        }
        return x;
    }
    bool merge(ll x, ll y) {
        x = root(x);
        y = root(y);
        if (x == y) return false;
        if (siz[x] < siz[y]) swap(x, y);
        siz[x] += siz[y];
        par[y] = x;
        return true;
    }

    bool issame(ll x, ll y) {
        return root(x) == root(y);
    }

    ll size(ll x) {
        return siz[root(x)];
    }
};

long long modpow(long long a, long long n, long long mod) {
    if (n < 0)return 0;
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

long long modinv(long long a, long long mod) {
    return modpow(a, mod - 2, mod);
}

vector<int> tpsort(vector<vector<int>>& G) {
    int V = G.size();
    vector<int> sorted_vertices;
    queue<int> que;
    vector<int> indegree(V);
    for (int i = 0; i < V; i++) {
        for (int j = 0; j < G[i].size(); j++) {
            indegree[G[i][j]]++;
        }
    }
    for (int i = 0; i < V; i++) {
        if (indegree[i] == 0) {
            que.push(i);
        }
    }
    while (que.empty() == false) {
        int v = que.front();
        que.pop();
        for (int i = 0; i < G[v].size(); i++) {
            int u = G[v][i];
            indegree[u] -= 1;
            if (indegree[u] == 0) que.push(u);
        }
        sorted_vertices.push_back(v);
    }
    return sorted_vertices;
}

struct Point
{
    double x;
    double y;
};

struct LineSegment
{
    Point start;
    Point end;
};

double tenkyori(const LineSegment& line, const Point& point)
{
    double x0 = point.x, y0 = point.y;
    double x1 = line.start.x, y1 = line.start.y;
    double x2 = line.end.x, y2 = line.end.y;

    double a = x2 - x1;
    double b = y2 - y1;
    double a2 = a * a;
    double b2 = b * b;
    double r2 = a2 + b2;
    double tt = -(a * (x1 - x0) + b * (y1 - y0));

    if (tt < 0)
        return sqrt((x1 - x0) * (x1 - x0) + (y1 - y0) * (y1 - y0));
    else if (tt > r2)
        return sqrt((x2 - x0) * (x2 - x0) + (y2 - y0) * (y2 - y0));

    double f1 = a * (y1 - y0) - b * (x1 - x0);
    return sqrt((f1 * f1) / r2);
}

ll merge_cnt(vector<int>& a) {
    int n = a.size();
    if (n <= 1) { return 0; }

    ll cnt = 0;
    vector<int> b(a.begin(), a.begin() + n / 2);
    vector<int> c(a.begin() + n / 2, a.end());

    cnt += merge_cnt(b);
    cnt += merge_cnt(c);

    int ai = 0, bi = 0, ci = 0;
    while (ai < n) {
        if (bi < b.size() && (ci == c.size() || b[bi] <= c[ci])) {
            a[ai++] = b[bi++];
        }
        else {
            cnt += n / 2 - bi;
            a[ai++] = c[ci++];
        }
    }
    return cnt;
}

struct edge { ll to, cost; };
typedef pair<ll, ll> P;
struct graph {
    ll V;
    vector<vector<edge> > G;
    vector<ll> d;

    graph(ll n) {
        init(n);
    }

    void init(ll n) {
        V = n;
        G.resize(V);
        d.resize(V);
        for (int i = 0; i < V; i++) {
            d[i] = 2000000000000000000;
        }
    }

    void add_edge(ll s, ll t, ll cost) {
        edge e;
        e.to = t, e.cost = cost;
        G[s].push_back(e);
    }

    void dijkstra(ll s) {
        for (int i = 0; i < V; i++) {
            d[i] = 2000000000000000000;
        }
        d[s] = 0;
        priority_queue<P, vector<P>, greater<P> > que;
        que.push(P(0, s));
        while (!que.empty()) {
            P p = que.top(); que.pop();
            ll v = p.second;
            if (d[v] < p.first) continue;
            for (auto e : G[v]) {
                if (d[e.to] > d[v] + e.cost) {
                    d[e.to] = d[v] + e.cost;
                    que.push(P(d[e.to], e.to));
                }
            }
        }
    }
};


int main() {
    ll n;
    cin >> n;
    vector<ll> z(n);
    ll k = 0;
    ll ans = 0;
    for (int i = 0; i < n; i++) {
        cin >> z[i];
        k = max(k, z[i]);
        ans += k - z[i];
    }
    cout << ans << endl;
}