Problem: One day Kefa found n baloons. For convenience, we denote color of i-th baloon as si — lowercase letter of the Latin alphabet. Also Kefa has k friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 100) — the number of baloons and friends.

Next line contains string s — colors of baloons.

Output

Answer to the task — «YES» or «NO» in a single line.

You can choose the case (lower or upper) for each letter arbitrary.

Examples

Input

4 2
aabb


Output

YES


Input

6 3
aacaab


Output

NO

Note

In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.

In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».

Solution (in PYTHON3):

Solution:
from sys  import stdin,stdout

import bisect

import math

from collections import deque,Counter

mod=10**9 +7 

def st():
    return list(stdin.readline().strip())

def inp():
    return int(stdin.readline())

def li():
    return list(map(int,stdin.readline().split()))

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

def pr(n):
    stdout.write(str(n)+"\n")
    
def DFS(dictionary,vertex,visited):
    visited[vertex]=True
    stack=[vertex]
    print(vertex)
    while stack:
        a=stack.pop()
        for i in dictionary[a]:
            if not visited[i]:
                print(i)
                visited[i]=True
                stack.append(i)


def BFS(dictionary, vertex,visited):
    visited[vertex]=True
    q=deque()
    q.append(vertex)
    while q:
        a=q.popleft()
        for i in dictionary[a]:
            if not visited[i]:
                visited[i]=True
                q.append(i)
                print(i)

                


def soe(limit):
    l=[1]*(limit+1)
    l[0]=0
    l[1]=0
    prime=[]
    for i in range(2,limit+1):
        if l[i]:
            for j in range(i*i,limit+1,i):
                l[j]=0
    
    for i in range(2,limit+1):
        if l[i]:
            prime.append(i)
    return prime

def segsoe(low,high):
    limit=int(high**0.5)+1
    prime=soe(limit)
    n=high-low+1
    l=[0]*(n+1)
    for i in range(len(prime)):
        lowlimit=(low//prime[i])*prime[i]
        if lowlimit<low:
            lowlimit+=prime[i]
        if lowlimit==prime[i]:
            lowlimit+=prime[i]
        for j in range(lowlimit,high+1,prime[i]):
            l[j-low]=1
    for i in range(low,high+1):
        if not l[i-low]:
            if i!=1:
                print(i)
                
def gcd(a,b):
    while b:
        a=a%b
        b,a=a,b
    return a

def power(a,n):
    r=1
    while n:
        if n&1:
            r=(r*a)
        a*=a
        n=n>>1
    return r

def issum(l,x):
    c=0
    i=0
    j=len(l)-1
    while i<j:
        a=l[i]+l[j]
        if a>x:
            j-=1
        elif a<x:
            i+=1
        else:
            c+=1
            i+=1
            j-=1
    return c
    
    
def solve():
    c=0
    n,k=mp()
    s=input()
    d=Counter(s)
    for i in d:
        if d[i]>k:
            pr("NO")
            return
    pr("YES")

for _ in range(1):
    solve()