Problem: You have a playlist consisting of n songs. The i-th song is characterized by two numbers t_i and b_i — its length and beauty respectively. The pleasure of listening to set of songs is equal to the total length of the songs in the set multiplied by the minimum beauty among them. For example, the pleasure of listening to a set of 3 songs having lengths [5, 7, 4] and beauty values [11, 14, 6] is equal to (5 + 7 + 4) ⋅ 6 = 96.

You need to choose at most k songs from your playlist, so the pleasure of listening to the set of these songs them is maximum possible.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3 ⋅ 10^5) – the number of songs in the playlist and the maximum number of songs you can choose, respectively.

Each of the next n lines contains two integers t_i and b_i (1 ≤ t_i, b_i ≤ 10^6) — the length and beauty of i-th song.

Output

Print one integer — the maximum pleasure you can get.

Examples

Input


4 3
4 7
15 1
3 6
6 8


Output


78


Input


5 3
12 31
112 4
100 100
13 55
55 50


Output


10000

Note

In the first test case we can choose songs {1, 3, 4}, so the total pleasure is (4 + 3 + 6) ⋅ 6 = 78.

In the second test case we can choose song 3. The total pleasure will be equal to 100 ⋅ 100 = 10000.

Solution (in PYTHON3):

Solution:
from copy import deepcopy

class PriorityQueue():
    def __init__(self):
        self.elements = [(0, None)]
        self.size = 0
        self.capacity = 1000

    def push(self, item):
        # the first element of item is the index, it must be digits
        self.size += 1
        self.elements.append(item)
        i = self.size
        
        while self.elements[i//2][0]>item[0]:
            self.elements[i] = self.elements[i//2]
            i = i // 2
        self.elements[i] = item
    def poll(self):
        if self.size <= 0:
            return None
        min_element = deepcopy(self.elements[1])
        last_element = self.elements.pop()
        self.size -= 1
        i = 1
        while (i*2)<=self.size:
            child = i * 2 # left child
            if (child != self.size and (self.elements[child+1][0] < self.elements[child][0])):
                child += 1
            if last_element[0] > self.elements[child][0]:
                self.elements[i] = self.elements[child]
            else:
                break
            
            i = child
        if self.size > 0:
            self.elements[i] = last_element
        return min_element




import heapq


n,k = tuple(map(int, input().split(' ')))
s = [] # it's possible to have the same value of beaty
for i in range(n):
    t = tuple(map(int, input().split(' ')))
    s.append(t)


def get_beauty(e):
    return e[1]
s.sort(reverse=True, key=get_beauty)
max_score = 0
sum_len = 0
heap = []
try:
    for i in range(n):
        length = s[i][0]
        beauty = s[i][1]
        sum_len += length
        if i+1 > k:
            sum_len-=heapq.heappop(heap) # 
        heapq.heappush(heap,length) # push before pop?
        score = (sum_len) * beauty
        max_score = max(max_score, score)
except Exception as e:
    print(e)
    

print(max_score)