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 xxtill 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
3 10 4 6 7Output
8Input
2 12 1 10Output
9Input
2 7 3 4Output
6Note
In the first example, one of possible optimal solutions is to insert value x=3x=3before 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 a1a1and 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.
题解：
写出状态方程，然后预处理：
想象插入数字其实是将一个数字拆成两个数字（两个操作），那么：
开灯时长为：在此之前的开灯时长和在此之后的关灯时长加上被拆成两个数字区间的变化（这个区间变化要考虑原区间是开着的还是关着的）；第i次操作时亮灯时长记为b[i]。如果本区间为开灯状态且区间长度大于1，则操作后开灯时长为：b[i]+m-a[i]-(b[n+1]-b[i])-1，如果本区间本来为关灯状态且区间长度大于1，则则操作后开灯时长为：b[i]+a[i]-a[i-1]-1+m-a[i]-(b[n+1]-b[i])。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
using namespace std;
int n;
const int maxn=1e5+10;
typedef long long ll;
ll m;
ll a[maxn],b[maxn];
int main()
{
bool f=1;
scanf("%d%lld",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%lld",&a[i]);
b[i]=b[i-1]+f*(a[i]-a[i-1]);
f=1-f;
}
a[n+1]=m;
b[n+1]=b[n]+f*(m-a[n]);
ll ans=b[n+1];
ll mymax;
for(int i=1;i<=n+1;i++)
{
if(a[i]-a[i-1]>1)
{
if(i&1){
mymax=b[i]*2+m-1-a[i]-b[n+1];
ans=max(ans,mymax);
}
else
{
mymax=2*b[i]-a[i-1]-1+m-b[n+1];
ans=max(ans,mymax);
}
}
}
printf("%lld\n",ans);
return 0;
}