You have N cards with different numbers on them. Your goal is to find a card with a maximal number. At the beginning all cards are put into the hat. You start getting them one by one and look at the numbers on them. After each card you can select it and stop the process. If it is really the card with the maximal number you win otherwise you lose. Also you can skip the current card and continue process. Fortunately you have a friend who helps with a good strategy: you pull X cards and memorize their values. Then you continue the process and select as answer the first card with value greater than the maximal value you memorized. Unfortunately you don't know the value of Xthat maximizes you chances of winning. Your task is to find X.

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <queue>

#include <map>

#include <set>

#include <vector>

#include <string>

#include <stack>

#include <bitset>

#define INF 1000000005

#define eps 1e-10

#define PI acos(-1.0)

#define K (0.017453292519943295769236907684886l)

#define LL long long

#define ULL unsigned long long

using namespace std;

const int maxn = 100005;

int a[maxn], n, m;

struct Node

{

    int val, pos;

}B[maxn];

bool cmp(const Node &x, const Node &y)

{

    return x.val < y.val;

}

LL Bel[maxn], Num[maxn], Prod[maxn];

int Get(int x)

{

    if (x > B[m].val) return m;

    int l = 1, r = m, pos = 0;

    while(l <= r)

    {

        int mid = (l + r) / 2;

        if (x >= B[mid].val)

        {

            pos = mid; l = mid + 1;

        }

        else r = mid - 1;

    }

    return pos;

}

int main()

{

    LL ans = 0;

    scanf("%d%d", &n, &m);

    for (int i = 1; i <= n; i++)

        scanf("%d", &a[i]);

    for (int i = 1; i <= m; i++)

        {

            scanf("%d", &B[i].val);

            B[i].pos = i;

        }

    sort(B + 1, B + 1 + m, cmp);

    for (int i = 1; i <= m; i++)

    {

        Bel[i] = Bel[i - 1] + B[i].pos;

        Num[i] = Num[i - 1] + B[i].val;

        Prod[i] = Prod[i - 1] + 1LL * B[i].pos * B[i].val;

      //  printf("%I64d %I64d %I64d\n", Bel[i], Num[i], Prod[i]);

    }

    for (int i = 1; i <= n; i++)

    {

        int pos = Get(a[i]);

     //   printf("%d\n", pos);

        ans = ans + 1LL * pos * i * a[i] + Prod[pos];

        ans = ans - 1LL * a[i] * Bel[pos] - 1LL * i * Num[pos];

        //

      // if (i == 1) printf("%I64d\n", ans);

        LL temp = 1LL * (m - pos) * i * a[i] + Prod[m] - Prod[pos];

        temp = temp - 1LL * a[i] * (Bel[m] - Bel[pos]) - 1LL * i * (Num[m] - Num[pos]);

        ans -= temp;

      //  if (i == 1) printf("%I64d\n", ans);

    }

    printf("%I64d\n", ans);

    return 0;

}