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.

Source: codeforces | Difficulty: VERY_HARD
Tags: Number theory

Mathematics

Complete search


Solutions:
Primes = [2, 3, 5, 7]

def check(a, b):
	for pr in Primes:
		if a % pr == 0 and b % pr == 0:
			return False
	while b:
		a %= b
		a ^= b
		b ^= a
		a ^= b
	return a == 1
(n, l, r) = map(int, input().split())
if n == 1:
	ans = r - l + 1
elif n == 2:
	ans = (r - l + 1) * (r - l)
else:
	n -= 1
	l -= 1
	ans = 0
	for q in range(2, 100000):
		upper = r // q ** n
		if upper == 0:
			break
		for p in range(1, q):
			if check(q, p):
				ans += max(0, upper - l // p ** n)
	ans *= 2
print(ans)


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
		bound = int(r ** (1 / (n - 1)))
		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))