Light It Up CF1000B 思维

Light It Up

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment 00 and turn power off at moment MM. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp.

The lamp allows only good programs. Good program can be represented as a non-empty array aa, where 0<a1<a2<⋯<a|a|<M0<a1<a2<⋯<a|a|<M. All aiai must be integers. Of course, preinstalled program is a good program.

The lamp follows program aa in next manner: at moment 00 turns power and light on. Then at moment aiai the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment 11 and then do nothing, the total time when the lamp is lit will be 11. Finally, at moment MM the lamp is turning its power off regardless of its state.

Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program aa, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of aa, or even at the begining or at the end of aa.

Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from xx till moment yy, then its lit for y−xy−x units of time. Segments of time when the lamp is lit are summed up.

Input

First line contains two space separated integers nn and MM (1≤n≤1051≤n≤105, 2≤M≤1092≤M≤109) — the length of program aa and the moment when power turns off.

Second line contains nn space separated integers a1,a2,…,ana1,a2,…,an (0<a1<a2<⋯<an<M0<a1<a2<⋯<an<M) — initially installed program aa.

Output

Print the only integer — maximum possible total time when the lamp is lit.

Examples

input

Copy

3 10
4 6 7

output

Copy

8

input

Copy

2 12
1 10

output

Copy

9

input

Copy

2 7
3 4

output

Copy

6

Note

In the first example, one of possible optimal solutions is to insert value x=3x=3 before a1a1, so program will be [3,4,6,7][3,4,6,7] and time of lamp being lit equals (3−0)+(6−4)+(10−7)=8(3−0)+(6−4)+(10−7)=8. Other possible solution is to insert x=5x=5 in appropriate place.

In the second example, there is only one optimal solution: to insert x=2x=2 between a1a1 and a2a2. Program will become [1,2,10][1,2,10], and answer will be (1−0)+(10−2)=9(1−0)+(10−2)=9.

In the third example, optimal answer is to leave program untouched, so answer will be (3−0)+(7−4)=6(3−0)+(7−4)=6.

题意：在0-m间插入数字使得相邻两数相差值的和最大

分析：要使插入的数能够让差值和最大则插入的位置肯定是每个数减一的位置，接下来在插入的时候求个前缀和，后缀和就行了

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e5 + ;
const int mod = 1e9 + ;
typedef long long ll;
ll a[maxn], n, m;
int main(){
    std::ios::sync_with_stdio(false);
    while( cin >> n >> m ) {
        ll ans = , on = , off = ;
        for( ll i = ; i <= n; i ++ ) {
            cin >> a[i];
        }
        a[] = , a[n+] = m;
        for( ll i = n; i >= ; i -- ) {
            if( i& ) {
                off += a[i+] - a[i] - ;
            } else {
                on += a[i+] - a[i] - ;
            }
            if( a[i+] != a[i] +  ) {
                ans = max( ans, off - on );
            }
            if( i& ) {
                off ++;
            } else {
                on ++;
            }
        }
        cout << ans + on << endl;
    }
    return ;
}