As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.

Elections are coming. You know the number of voters and the number of parties — n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.

The United Party of Berland has decided to perform a statistical study — you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.

Input

The first line of input contains two integers n and m (1 ≤ n, m ≤ 3000) — the number of voters and the number of parties respectively.

Each of the following n lines contains two integers p_i and c_i (1 ≤ p_i ≤ m, 1 ≤ c_i ≤ 10^9) — the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.

The United Party of Berland has the index 1.

Output

Print a single number — the minimum number of bytecoins needed for The United Party of Berland to win the elections.

Examples

Input

1 2
1 100


Output

0


Input

5 5
2 100
3 200
4 300
5 400
5 900


Output

500


Input

5 5
2 100
3 200
4 300
5 800
5 900


Output

600

Note

In the first sample, The United Party wins the elections even without buying extra votes.

In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.

In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**7)
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
import math  
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l
p={}
q=[]
def dfs(n,d,v):
    global p
    global q
    v[n]=1
    x=d[n]
    q.append(n)
    j=0
    for i in x:
        if i not in v:
            f=dfs(i,d,v)
            p[i]=f
            j+=f
    return j+1

"""*******************************************************"""
def main():
    n,m=cin()
    x=0
    a=[]
    d={}
    for i in range(n):
        j,k=cin()
        if(j==1):
            x+=1
        else:
            a.append((k,j))
            if j in d:
                d[j].append(k)
            else:
                d[j]=[k]
    # n=len(a)
    a.sort()
    for i in d:
        d[i].sort()
    # print(d)
    an=1000000000000
    for i in range(max(0,x-1),n):
        ans=0
        r=x
        l={}
        for j in d:
            p=len(d[j])-i
            # print(j,p)
            for k in range(0,p):
                ans+=d[j][k]
                l[(d[j][k],j)]=1
            r+=max(0,p)
        r=i+1-r
        # print(r,ans)
        f=0
        for j in a:
            if(f>=r):
                break
            if j in l and l[j]>0:
                l[j]-=1
            else:
                f+=1
                ans+=j[0]
            # print(a,l)
        # print(ans)
        an=min(an,ans)
    if(an==1000000000000):
        print(0)
    else:
        print(an)
######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A

# threading.Thread(target=main).start()
if __name__== "__main__":
  main()