Valera loves his garden, where n fruit trees grow.

This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).

Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more than v fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?

Input

The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day. 

Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.

Output

Print a single integer — the maximum number of fruit that Valera can collect. 

Examples

Input

2 3
1 5
2 3


Output

8


Input

5 10
3 20
2 20
1 20
4 20
5 20


Output

60

Note

In the first sample, in order to obtain the optimal answer, you should act as follows. 

  * On the first day collect 3 fruits from the 1-st tree. 
  * On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree. 
  * On the third day collect the remaining fruits from the 2-nd tree. 



In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.
import sys
from collections import defaultdict, Counter
from itertools import permutations, combinations
from math import sin, cos, asin, acos, tan, atan, pi

sys.setrecursionlimit(10 ** 6)

def pyes_no(condition, yes = "YES", no = "NO", none = "-1") :
  if condition == None:
    print (none)
  elif condition :
    print (yes)
  else :
    print (no)

def plist(a, s = ' ') :
  print (s.join(map(str, a)))

def rint() :
  return int(sys.stdin.readline())

def rstr() :
  return sys.stdin.readline().strip()


def rints() :
  return map(int, sys.stdin.readline().split())

def rfield(n, m = None) :
  if m == None :
    m = n
  
  field = []
  for i in xrange(n) :
    chars = sys.stdin.readline().strip()
    assert(len(chars) == m)
    field.append(chars)
  return field

def pfield(field, separator = '') :
  print ('\n'.join(map(lambda x: separator.join(x), field)))

def check_field_equal(field, i, j, value) :
  if i >= 0 and i < len(field) and j >= 0 and j < len(field[i]) :
    return value == field[i][j]
  return None 

def digits(x, p) :
  digits = []
  while x > 0 :
    digits.append(x % p)
    x //= p
  return digits[::-1]

def undigits(x, p) :
  value = 0
  for d in x :
    value *= p
    value += d
  return value

def modpower(a, n, mod) :
  r = a ** (n % 2)
  if n > 1 :
    r *= modpower(a, n // 2, mod) ** 2
  return r % mod

def gcd(a, b) :
  if a > b :
    a, b = b, a
  
  while a > 0 :
    a, b = b % a, a

  return b

def vector_distance(a, b) :
  diff = vector_diff(a, b)
  
  return scalar_product(diff, diff) ** 0.5

def vector_inverse(v) :
  r = [-x for x in v]

  return tuple(r)

def vector_diff(a, b) :
  return vector_sum(a, vector_inverse(b))

def vector_sum(a, b) :
  r = [c1 + c2 for c1, c2 in zip(a, b)]
    
  return tuple(r)

def scalar_product(a, b) :
  r = 0
  for c1, c2 in zip(a, b) :
    r += c1 * c2

  return r

def check_rectangle(points) :
  assert(len(points) == 4)

  A, B, C, D = points

  for A1, A2, A3, A4 in [
    (A, B, C, D),
    (A, C, B, D),
    (A, B, D, C),
    (A, C, D, B),
    (A, D, B, C),
    (A, D, C, B),
  ] :
    sides = (
      vector_diff(A1, A2),
      vector_diff(A2, A3),
      vector_diff(A3, A4),
      vector_diff(A4, A1),
    )
    if all(scalar_product(s1, s2) == 0 for s1, s2 in zip(sides, sides[1:])) :
       return True
  return False

def check_square(points) :
  if not check_rectangle(points) :
    return False
  A, B, C, D = points

  for A1, A2, A3, A4 in [
    (A, B, C, D),
    (A, C, B, D),
    (A, B, D, C),
    (A, C, D, B),
    (A, D, B, C),
    (A, D, C, B),
  ] :
    side_lengths = [
      (first[0] - next[0]) ** 2 + (first[1] - next[1]) ** 2 for first, next in zip([A1, A2, A3, A4], [A2, A3, A4, A1])
    ]
    if len(set(side_lengths)) == 1 :
      return True
    
  return False

def check_right(p) :
  # Check if there are same points
  for a, b in [
    (p[0], p[1]),
    (p[0], p[2]),
    (p[1], p[2]),
  ] :
    if a[0] == b[0] and a[1] == b[1] :
      return False

  a, b, c = p
  a, b, c = vector_diff(a, b), vector_diff(b, c), vector_diff(c, a)   

  return scalar_product(a, b) * scalar_product(a, c) * scalar_product(b, c) == 0

def modmatrixproduct(a, b, mod) :
  n, m1 = len(a), len(a[0])
  m2, k = len(b), len(b[0])

  assert(m1 == m2)
  m = m1

  r = [[0] * k for i in range(n)]
  for i in range(n) :
    for j in range(k) :
      for l in range(m) :
        r[i][j] += a[i][l] * b[l][j]
      r[i][j] %= mod
  return r

def modmatrixpower(a, n, mod) :
  magic = 2
  for m in [2, 3, 5, 7] :
    if n % m == 0 :
      magic = m
      break

  r = None
  if n < magic : 
    r = a
    n -= 1
  else :
    s = modmatrixpower(a, n // magic, mod)
    r = s
    for i in range(magic - 1) :
      r = modmatrixproduct(r, s, mod)

  for i in range(n % magic) : 
    r = modmatrixproduct(r, a, mod)
  
  return r

n, v = rints()

fruits = [0] * 3002
 
for i in range(n) :
  day, count = rints()
  fruits[day] += count

prev_count = 0
overall = 0
for i in range(3002) :
  count = fruits[i]
  today = v

  old = min(prev_count, today)
  today -= old
  overall += old

  current = min(count, today)
  prev_count = count - current
  overall += current

print overall
