Problem: For given n, l and r find the number of distinct geometrical progression, each of which contains n distinct integers not less than l and not greater than r. In other words, for each progression the following must hold: l ≤ ai ≤ r and ai ≠ aj , where a1, a2, ..., an is the geometrical progression, 1 ≤ i, j ≤ n and i ≠ j.

Geometrical progression is a sequence of numbers a1, a2, ..., an where each term after first is found by multiplying the previous one by a fixed non-zero number d called the common ratio. Note that in our task d may be non-integer. For example in progression 4, 6, 9, common ratio is <image>.

Two progressions a1, a2, ..., an and b1, b2, ..., bn are considered different, if there is such i (1 ≤ i ≤ n) that ai ≠ bi.

Input

The first and the only line cotains three integers n, l and r (1 ≤ n ≤ 107, 1 ≤ l ≤ r ≤ 107).

Output

Print the integer K — is the answer to the problem.

Examples

Input

1 1 10


Output

10

Input

2 6 9


Output

12

Input

3 1 10


Output

8

Input

3 3 10


Output

2

Note

These are possible progressions for the first test of examples: 

  * 1; 
  * 2; 
  * 3; 
  * 4; 
  * 5; 
  * 6; 
  * 7; 
  * 8; 
  * 9; 
  * 10. 



These are possible progressions for the second test of examples: 

  * 6, 7; 
  * 6, 8; 
  * 6, 9; 
  * 7, 6; 
  * 7, 8; 
  * 7, 9; 
  * 8, 6; 
  * 8, 7; 
  * 8, 9; 
  * 9, 6; 
  * 9, 7; 
  * 9, 8. 



These are possible progressions for the third test of examples: 

  * 1, 2, 4; 
  * 1, 3, 9; 
  * 2, 4, 8; 
  * 4, 2, 1; 
  * 4, 6, 9; 
  * 8, 4, 2; 
  * 9, 3, 1; 
  * 9, 6, 4. 



These are possible progressions for the fourth test of examples: 

  * 4, 6, 9; 
  * 9, 6, 4. 

Solution (in PYTHON3):

Solution:
import math

small=[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523]





smallest=[0 for i in range(4001)]
smallest[0]=1
smallest[1]=1
for i in range(2,len(smallest)):
    if smallest[i]==0:
        for j in range(2*i,len(smallest),i):
            smallest[j]=i
        smallest[i]=i
def listfactor(N):
    
    if N==2:
        return [2]
    elif N==3:
        return [3]
    else:
        temp=[]
        while(N>1):
            t=smallest[N]
            temp+=[t]
            N=N//t
        temp=temp[::-1]    
        ANS=[temp[0]]
        cur=temp[0]
        for i in range(len(temp)):
            if temp[i]>cur:
                cur=temp[i]
                ANS+=[cur]
        return ANS        

def check(a,b):
    for x in small:
        if (a%x==0 and b%x==0)==True:
            return 0
        if x>max(a,b)**0.5:
            return 1
    return 1


def ceil(x):
    if x!=int(x):
        return int(x)+1
    else:
        return int(x)

def gcd(a,b):
    a,b=min(a,b),max(a,b)
    if b%a==0:
        return a
    else:
        return gcd(a,b%a)


def find(n,l,r):
    if r-l+1<n:
        return 0
    elif n==1:
        return r-l+1
    elif n==2:
        return (r-l+1)*(r-l)
    elif n>=30:
        ans=0
        temp=0
        for d in range(2,10**10):
            temp=max(r//(d**(n-1))-l+1,0)
            ans+=temp
            if temp==0:
                break
        return 2*ans
    else:
        ans=0
        for d in range(2,10**10):
            temp=max(r//(d**(n-1))-l+1,0)
            ans+=temp
            if temp==0:
                break  
        #print(ans,'ans')        
        bound=int(r**(1/(n-1)))
        #print(bound,'bound')
        #q<=bound
        ANS=0
        
        
        for q in range(2,bound+1):
            CUR=listfactor(q)
            for p in range(q+1,bound+2):
                check=1
                for cc in CUR:
                    if p%cc==0:
                        check=0
                
                
                if check==1:
                    ANS+=max(r//(p**(n-1))-ceil(l/(q**(n-1)))+1,0)
        return 2*(ans+ANS)            
                    
                   
n,l,r=list(map(int,input().strip().split(' ')))    
print(find(n,l,r))