#!/bin/python3

import os
from collections import Counter
from typing import Literal

#
# Complete the 'is_valid' function below.
#
# The function is expected to return a STRING.
# The function accepts STRING s as parameter.
#


def is_valid(s: str) -> Literal["NO", "YES"]:
    """Given a string, determine if it is valid."""
    y = "YES"
    if len(s) == 1:  # optional
        return y
    match len(counts := {*(frequencies := Counter(s).values())}):
        case 1:
            return y
        case 2:
            meta_freq = Counter(frequencies)
            lo, hi = sorted(counts)
            if 1 == lo == meta_freq[lo]:
                return y
            if 1 == hi - lo == meta_freq[hi]:
                return y
    return "NO"


if __name__ == "__main__":
    result = is_valid(input())
    with open(os.environ["OUTPUT_PATH"], "w") as fptr:
        fptr.write(f"{result}\n")

#python 3
def isValid(s):
    keys = set(s)
    frequencies = {key:0 for key in keys}
    for letter in s:
        frequencies[letter] += 1
        
    counts = Counter(frequencies.values())
    maxCount = max(counts.values())
    mode = min([key for key, value in counts.items() if value == maxCount])
    diff = 0
    
    for letter,num in frequencies.items():
        if num != mode:
            if (abs(num - mode) == 1 or num == 1) and diff == 0:
                diff += 1
            else:
                return "NO"
    return "YES"

def isValid(s):
    # Write your code here
    stringDict={}
    
    for letter in s:
        stringDict[letter]= stringDict.get(letter,0) + 1 
    dictValues= list(stringDict.values())
    isRemovedOne=False
    for i in range(1,len(dictValues)):
        diff= abs(dictValues[i]- dictValues[i-1])
        if  diff == 0:
            continue
        else:
            if not isRemovedOne:
                isRemovedOne=True
                dictValues[i]-=1
            else:
                return "NO"
    return "YES"

def isValid(s):
    cts=Counter(s) #Count occurences of string
    ct2=Counter(cts.values()) #Count unique occurences of string occurences
    #if only one item in ct2 then all occurences are equal, 
    #if two items then check if diff is 1 and second most has a count of 1
    #Special case: if the second occurences of occurences equals 1, 
    #it means the second 1 occurrence can be removed to make a valid string.
    kc=ct2.most_common()
    return 'YES' if len(kc)==1 or (len(kc)==2 and \
     (((kc[1][0]-kc[0][0])==1 or kc[1][0]==1) and kc[1][1]==1)) else 'NO'

def isValid(s):
    # Write your code here
    holder = {}
    for char in s:
        if char in holder:
            holder[char] += 1
        else:
            holder[char] = 1
    val = list(holder.values())
    val.sort()
    test1 = set(val[1:])
    test2 = set(val[:-1])
    test2.add(val[-1]-1)
    result = "YES" if len(test1) == 1 or len(test2) == 1 else "NO"
    return result