B. Light It Up 思维题

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 aaand 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.

这题主要运用了一个思想 就是加入一个点就相当于把一个点拆成了两个 ，

根据题目的要求贪心 ，将a[i] 拆成两个点的时候 就是 拆成a[i]-1 和 a[i]；这个我们就可以获得最优解 ，

ans[i]记录的题目要求值得前缀和

因为改变了a[i] 所以就需要a[i]后的关灯的值

m - a[i] - (ans[n + 1] - ans[i]) 这个为a[i]后关灯的值

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = 4e5 + ;
 typedef long long LL;
 int a[maxn], n, m, ans[maxn];
 int main() {
     cin >> n >> m;
     a[] = , a[n + ] = m;
     for (int i =  ; i <= n ; i++) scanf("%d", &a[i]);
     int flag = ;
     ans[] = ;
     for (int i =  ; i <= n +  ; i++) {
         ans[i] = ans[i - ] + flag * (a[i] - a[i - ]);
         flag ^= ;
     }
     int cnt = ans[n + ];
     for (int i =  ; i <= n +  ; i++) {
         cnt = max(cnt, ans[i] -  + m - a[i] - (ans[n + ] - ans[i]));
     }
     printf("%d\n", cnt);
     return ;
 }