human
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.

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.

Examples

Input

3 20
2 80
9 120
16 1

Output

191

Input

3 20
2 80
9 1
16 120

Output

192

Input

1 100000000000000
50000000000000 1

Output

0

Input

15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000...
Solve the task in PYTHON3.
 ... What is a code to teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter?
assistant
#Here is code to teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter.


class SegmentTree:
    def __init__(self, L, op, ide):
        self.op = op
        self.ide = ide
        self.sz = len(L)
        self.n = 1
        self.s = 1
        for i in range(1000):
            self.n *= 2
            self.s += 1
            if self.n >= self.sz:
                break
        self.node = [self.ide]*(2*self.n-1)

        for i in range(self.sz):
            self.node[i+self.n-1] = L[i]

        for i in range(self.n-2,-1,-1):
            self.node[i] = self.op(self.node[i*2+1],self.node[i*2+2])
    
    def add(self, a, x):
        k = a+self.n-1
        self.node[k] += x
        for i in range(1000):
            k = (k-1)//2
            self.node[k] = self.op(self.node[2*k+1], self.node[2*k+2])
            if k <= 0:
                break
    
    def substitute(self, a, x):
        k = a+self.n-1
        self.node[k] = x
        for i in range(1000):
            k = (k-1)//2
            self.node[k] = self.op(self.node[2*k+1], self.node[2*k+2])
            if k <= 0:
                break

    def get_one(self, a):
        k = a+self.n-1
        return self.node[k]
    
    def get(self, l, r):
        res = self.ide
        n = self.n
        if self.sz <= r or 0 > l:
            print("ERROR: the indice are wrong.")
            return False

        for i in range(self.s):
            count = 2**i-1
            a = (r+1)//n
            b = (l-1)//n
            if a-b == 3:
                res = self.op(self.node[count+b+1],res)
                res = self.op(self.node[count+b+2],res)
                right = a*n
                left = (b+1)*n-1
                break
            if a-b == 2:
                res = self.op(self.node[count+b+1],res)
                right = a*n
                left = (b+1)*n-1
                break
            n = n//2

        #left
        n1 = n//2        
        for j in range(i+1,self.s):
            count = 2**j-1 
            a = (left+1)//n1
            b = (l-1)//n1
            if a-b == 2:
                res =...