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 $\mathrm{00 and\; turn\; power\; off\; at\; moment MM.\; Moreover,\; the\; lamp\; allows\; you\; to\; set\; a\; program\; of\; switching\; its\; state\; (states\; are\; "lights\; on"\; and\; "lights\; off").\; Unfortunately,\; some\; program\; is\; already\; installed\; into\; the\; lamp.}$

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

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

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 $\mathrm{aa,\; so\; it\; still\; should\; be\; a good program\; after\; alteration.\; Insertion\; can\; be\; done\; between\; any\; pair\; of\; consecutive\; elements\; of aa,\; or\; even\; at\; the\; begining\; or\; at\; the\; end\; of aa.}$

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 $\mathrm{xx till\; moment yy,\; then\; its\; lit\; for y-xy-x units\; of\; time.\; Segments\; of\; time\; when\; the\; lamp\; is\; lit\; are\; summed\; up.}$

Input

First line contains two space separated integers $\mathrm{nn and MM (1\le n\le 1051\le n\le 105, 2\le M\le 1092\le M\le 109)\; —\; the\; length\; of\; program aaand\; the\; moment\; when\; power\; turns\; off.}$

Second line contains $\mathrm{nn space\; separated\; integers a1,a2,\dots ,ana1,a2,\dots ,an (0<a1<a2<\cdots <an<M0<a1<a2<\cdots <an<M)\; —\; initially\; installed\; program aa.}$

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 $\mathrm{x=3x=3 before a1a1,\; so\; program\; will\; be [3,4,6,7][3,4,6,7]and\; time\; of\; lamp\; being\; lit\; equals (3-0)+(6-4)+(10-7)=8(3-0)+(6-4)+(10-7)=8.\; Other\; possible\; solution\; is\; to\; insert x=5x=5 in\; appropriate\; place.}$

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

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 + 10;
typedef long long LL;
int a[maxn], n, m, ans[maxn];
int main() {
    cin >> n >> m;
    a[0] = 0, a[n + 1] = m;
    for (int i = 1 ; i <= n ; i++) scanf("%d", &a[i]);
    int flag = 1;
    ans[0] = 0;
    for (int i = 1 ; i <= n + 1 ; i++) {
        ans[i] = ans[i - 1] + flag * (a[i] - a[i - 1]);
        flag ^= 1;
    }
    int cnt = ans[n + 1];
    for (int i = 1 ; i <= n + 1 ; i++) {
        cnt = max(cnt, ans[i] - 1 + m - a[i] - (ans[n + 1] - ans[i]));
    }
    printf("%d\n", cnt);
    return 0;
}