Problem:
Score : 500 points

Problem Statement

"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.

Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.

Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.

Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.

Constraints

1 ≤ N ≤ 10^5

2 ≤ C ≤ 10^{14}

1 ≤ x_1 < x_2 < ... < x_N < C

1 ≤ v_i ≤ 10^9

All values in input are integers.

Subscores

300 points will be awarded for passing the test set satisfying N ≤ 100.

Input

Input is given from Standard Input in the following format:

N C
x_1 v_1
x_2 v_2
:
x_N v_N

Output

If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.

Sample Input 1

3 20
2 80
9 120
16 1

Sample Output 1

191

There are three sushi on the counter with a circumference of 20 meters. If he walks two meters clockwise from the initial place, he can eat a sushi of 80 kilocalories. If he walks seven more meters clockwise, he can eat a sushi of 120 kilocalories. If he leaves now, the total nutrition taken in is 200 kilocalories, and the total energy consumed is 9 kilocalories, thus he can take in 191 kilocalories on balance, which is the largest possible value.

Sample Input 2

3 20
2 80
9 1
16 120

Sample Output 2

192

The second and third sushi have been swapped. Again, if he walks two meters clockwise from the initial place, he can eat a sushi of 80 kilocalories. If he walks six more meters counterclockwise this time, he can eat a sushi of 120 kilocalories. If he leaves now, the total nutrition taken in is 200 kilocalories, and the total energy consumed is 8 kilocalories, thus he can take in 192 kilocalories on balance, which is the largest possible value.

Sample Input 3

1 100000000000000
50000000000000 1

Sample Output 3

0

Even though the only sushi is so far that it does not fit into a 32-bit integer, its nutritive value is low, thus he should immediately leave without doing anything.

Sample Input 4

15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000

Sample Output 4

6500000000

All these sample inputs above are included in the test set for the partial score.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.*;
import java.util.*;

/**
 * @author baito
 */
public class Main
{
    static StringBuilder sb = new StringBuilder();
    static FastScanner sc = new FastScanner(System.in);
    static int INF = 10000;
    static long MOD = 1000000007;
    static int[] y4 = {0, 1, 0, -1};
    static int[] x4 = {1, 0, -1, 0};
    static int[] y8 = {0, 1, 0, -1, -1, 1, 1, -1};
    static int[] x8 = {1, 0, -1, 0, 1, -1, 1, -1};
    static long[] F;//factorial
    static boolean[] isPrime;
    static int[] primes;

    static int N;
    static long C;
    static long[] x, v;

    public static void main(String[] args)
    {

        N = sc.nextInt();
        C = sc.nextLong();
        x = new long[N];
        v = new long[N];
        for (int i = 0; i < N; i++)
        {
            x[i] = sc.nextLong();
            v[i] = sc.nextLong();
        }
        //longを忘れるなオーバーフローするぞ
        //右回り、左回り(先に動く)について、各地点での最高カロリーを記録
        long [] maxAs = new long[N + 1];
        long[] maxBs = new long[N + 1];
        long[] Avalues = new long[N];
        long[] Bvalues = new long[N];


        maxAs[0] = 0;
        long ate = 0;
        for (int i = 1; i < N + 1; i++)
        {
            maxAs[i] = Math.max(ate + v[i - 1] - 2 * x[i - 1], maxAs[i - 1]);
            ate += v[i-1];
        }
        maxBs[0] = 0;
        ate = 0;
        for (int i = 1; i < N + 1; i++)
        {
            maxBs[i] = Math.max(ate + v[N - i] - 2 * (C - x[N - i]), maxBs[i - 1]);
            ate += v[N-i];
        }
        //
        Avalues[0] = v[0] - x[0];
        ate = v[0];
        for (int i = 1; i < N; i++)
        {
            Avalues[i] = ate + v[i] - x[i];
            ate += v[i];
        }
        Bvalues[0] = v[N - 1] - (C - x[N - 1]);
        ate = v[N - 1];
        for (int i = 1; i < N; i++)
        {
            Bvalues[i] = ate + v[(N - i - 1)] - (C - x[N - i-1]);
            ate += v[i];
        }
        //最終地点を全探索
        long res = 0;
        for (int bi = 0; bi < N; bi++)
        {
            res = Math.max(res, Bvalues[N-bi-1] + maxAs[bi]);
            //a==bのパターン。つまり折り返さない
            res = Math.max(res,Avalues[bi]);
        }
        for (int ai = 0; ai < N; ai++)
        {
            res = Math.max(res, Avalues[ai] + maxBs[N-ai-1]);
            //a==bのパターン。つまり折り返さない
            res = Math.max(res,Bvalues[ai]);
        }
        System.out.println(res);
    }

    //次の順列に書き換える、最大値ならfalseを返す
    public static boolean nextPermutation(int A[])
    {
        int len = A.length;
        int pos = len - 2;
        for (; pos >= 0; pos--)
        {
            if (A[pos] < A[pos + 1]) break;
        }
        if (pos == -1) return false;

        //posより大きい最小の数を二分探索
        int ok = pos + 1;
        int ng = len;
        while (Math.abs(ng - ok) > 1)
        {
            int mid = (ok + ng) / 2;
            if (A[mid] > A[pos]) ok = mid;
            else ng = mid;

        }

        swap(A, pos, ok);
        reverse(A, pos + 1, len - 1);


        return true;
    }

    //次の順列に書き換える、最小値ならfalseを返す
    public static boolean prevPermutation(int A[])
    {
        int len = A.length;
        int pos = len - 2;
        for (; pos >= 0; pos--)
        {
            if (A[pos] > A[pos + 1]) break;
        }
        if (pos == -1) return false;

        //posより小さい最大の数を二分探索
        int ok = pos + 1;
        int ng = len;
        while (Math.abs(ng - ok) > 1)
        {
            int mid = (ok + ng) / 2;
            if (A[mid] < A[pos]) ok = mid;
            else ng = mid;

        }

        swap(A, pos, ok);
        reverse(A, pos + 1, len - 1);


        return true;
    }

    //↓nCrをmod計算するために必要。　***factorial(N)を呼ぶ必要がある***
    static long ncr(int n, int r)
    {
        factorial(n);
        return F[n] / (F[n - r] * F[r]);
    }

    static long modNcr(int n, int r)
    {
        long result = F[n];
        result = result * modInv(F[n - r]) % MOD;
        result = result * modInv(F[r]) % MOD;
        return result;
    }

    static long modInv(long n)
    {
        return modPow(n, MOD - 2);
    }

    static void factorial(int n)
    {
        F = new long[n + 1];
        F[0] = F[1] = 1;
        for (int i = 2; i <= n; i++)
        {
            F[i] = (F[i - 1] * i) % MOD;
        }
    }

    static long modPow(long x, long n)
    {
        long res = 1L;
        while (n > 0)
        {
            if ((n & 1) == 1)
            {
                res = res * x % MOD;
            }
            x = x * x % MOD;
            n >>= 1;
        }
        return res;
    }

    //↑nCrをmod計算するために必要

    static int gcd(int n, int r)
    {
        return r == 0 ? n : gcd(r, n % r);
    }

    static long gcd(long n, long r)
    {
        return r == 0 ? n : gcd(r, n % r);
    }

    static <T> void swap(T[] x, int i, int j)
    {
        T t = x[i];
        x[i] = x[j];
        x[j] = t;
    }

    static void swap(int[] x, int i, int j)
    {
        int t = x[i];
        x[i] = x[j];
        x[j] = t;
    }

    public static void reverse(int[] x)
    {
        int l = 0;
        int r = x.length - 1;
        while (l < r)
        {
            int temp = x[l];
            x[l] = x[r];
            x[r] = temp;
            l++;
            r--;
        }
    }

    public static void reverse(int[] x, int s, int e)
    {
        int l = s;
        int r = e;
        while (l < r)
        {
            int temp = x[l];
            x[l] = x[r];
            x[r] = temp;
            l++;
            r--;
        }
    }

    static int length(int a)
    {
        int cou = 0;
        while (a != 0)
        {
            a /= 10;
            cou++;
        }
        return cou;
    }

    static int length(long a)
    {
        int cou = 0;
        while (a != 0)
        {
            a /= 10;
            cou++;
        }
        return cou;
    }

    static int countC2(char[][] a, char c)
    {
        int co = 0;
        for (int i = 0; i < a.length; i++)
            for (int j = 0; j < a[0].length; j++)
                if (a[i][j] == c) co++;
        return co;
    }

    static void fill(int[][] a, int v)
    {
        for (int i = 0; i < a.length; i++)
            for (int j = 0; j < a[0].length; j++)
                a[i][j] = v;
    }

    static int max(int a, int b, int c)
    {
        return Math.max(a, Math.max(b, c));
    }

    static int abs(int a)
    {
        return Math.abs(a);
    }

    static class FastScanner
    {

        private BufferedReader reader = null;
        private StringTokenizer tokenizer = null;

        public FastScanner(InputStream in)
        {
            reader = new BufferedReader(new InputStreamReader(in));
            tokenizer = null;
        }

        public String next()
        {
            if (tokenizer == null || !tokenizer.hasMoreTokens())
            {
                try
                {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e)
                {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        /*public String nextChar(){
            return (char)next()[0];
        }*/
        public String nextLine()
        {
            if (tokenizer == null || !tokenizer.hasMoreTokens())
            {
                try
                {
                    return reader.readLine();
                } catch (IOException e)
                {
                    throw new RuntimeException(e);
                }
            }

            return tokenizer.nextToken("\n");
        }

        public long nextLong()
        {
            return Long.parseLong(next());
        }

        public int nextInt()
        {
            return Integer.parseInt(next());
        }

        public double nextDouble()
        {
            return Double.parseDouble(next());
        }

        public int[] nextIntArray(int n)
        {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
            {
                a[i] = nextInt();
            }
            return a;
        }

        public int[][] nextIntArray2(int h, int w)
        {
            int[][] a = new int[h][w];
            for (int hi = 0; hi < h; hi++)
            {
                for (int wi = 0; wi < w; wi++)
                {
                    a[hi][wi] = nextInt();
                }
            }
            return a;
        }

        public int[] nextIntArray21(int n, int scalar)
        {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt() * scalar + nextInt();
            return a;
        }

        public Integer[] nextIntegerArray(int n)
        {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
            {
                a[i] = nextInt();
            }
            return a;
        }

        public char[] nextCharArray(int n)
        {
            char[] a = next().toCharArray();

            return a;
        }

        public char[][] nextCharArray2(int h, int w)
        {
            char[][] a = new char[h][w];
            for (int i = 0; i < h; i++)
            {
                a[i] = next().toCharArray();
            }
            return a;
        }

        //スペースが入っている場合
        public char[][] nextCharArray2s(int h, int w)
        {
            char[][] a = new char[h][w];
            for (int i = 0; i < h; i++)
            {
                a[i] = nextLine().replace(" ", "").toCharArray();
            }
            return a;
        }

        public char[][] nextWrapCharArray2(int h, int w, char c)
        {
            char[][] a = new char[h + 2][w + 2];
            //char c = '*';
            int i;
            for (i = 0; i < w + 2; i++)
                a[0][i] = c;
            for (i = 1; i < h + 1; i++)
            {
                a[i] = (c + next() + c).toCharArray();
            }
            for (i = 0; i < w + 2; i++)
                a[h + 1][i] = c;
            return a;
        }

        //スペースが入ってる時用
        public char[][] nextWrapCharArray2s(int h, int w, char c)
        {
            char[][] a = new char[h + 2][w + 2];
            //char c = '*';
            int i;
            for (i = 0; i < w + 2; i++)
                a[0][i] = c;
            for (i = 1; i < h + 1; i++)
            {
                a[i] = (c + nextLine().replace(" ", "") + c).toCharArray();
            }
            for (i = 0; i < w + 2; i++)
                a[h + 1][i] = c;
            return a;
        }

        public long[] nextLongArray(int n)
        {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
            {
                a[i] = nextLong();
            }
            return a;
        }

        public long[][] nextLongArray2(int h, int w)
        {
            long[][] a = new long[h][w];
            for (int hi = 0; hi < h; hi++)
            {
                for (int wi = 0; wi < h; wi++)
                {
                    a[h][w] = nextLong();
                }
            }
            return a;
        }
    }
}