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

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 $\mathrm{nn and MM (1\le n\le 1051\le n\le 105, 2\le M\le 1092\le M\le 109)\; —\; the\; length\; of\; program aa and\; the\; moment\; when\; power\; turns\; off.}$

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 $(3-0)+(7-4)=6(3-0)+(7-4)=6.$

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 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
ll a[maxn], n, m;
int main(){
    std::ios::sync_with_stdio(false);
    while( cin >> n >> m ) {
        ll ans = 0, on = 0, off = 0;
        for( ll i = 1; i <= n; i ++ ) {
            cin >> a[i];
        }
        a[0] = 0, a[n+1] = m;
        for( ll i = n; i >= 0; i -- ) {
            if( i&1 ) {
                off += a[i+1] - a[i] - 1;
            } else {
                on += a[i+1] - a[i] - 1;
            }
            if( a[i+1] != a[i] + 1 ) {
                ans = max( ans, off - on );
            }
            if( i&1 ) {
                off ++;
            } else {
                on ++;
            }
        }
        cout << ans + on << endl;
    }
    return 0;
}