human
As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256

Output

YES

Input

512

Output

NO

Note

In the first sample number .

In the second sample number 512 can not be represented as a sum of two triangular numbers.
 Write me a Python code to print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes)?
assistant
#Here is Python code to print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

# n = int(input())

# def isTriangular(n):
#     start = 1
#     end = n
#     ans = -1
#     while True:
#         # input()
#         k = (start + end) // 2
#         # print(start, end, k)
#         if k*(k+1)/2 == n:
#             ans = k
#         elif k*(k+1)/2 > n:
#             end = k
#         else:
#             start = k

#         if(start >= end-1 or ans != -1):
#             return True if ans != -1 else False
#             break
    
# # print(isTriangular(253))

# funky = False
# for i in range(1, (-n//2)*-1):
#     if(isTriangular(i) and isTriangular(n - i)):
#         print('YES')
#         funky = True
#         break
# if funky == False:
#     print('NO')

import math

n = int(input())
for i in range(1, n):
    z = i*(i+1)
    if z > n:
        break
    
    t = int(math.sqrt(2*n-z))
    t = t * (t+1)
    if(t == 2 * n -z):
        print('YES')
        quit()
print('NO')

#include <stdio.h>
#include <math.h>
# int main(){
# int n,t;
# 	scanf("%d",&n);
# 	for(int i=1;i<n;i++){
# 		int z=i*(i+1);
# 		if(n<z)break;
		
# 		t=(int)sqrt(2*n-z);
# 		t=t*(t+1);
# 		if(t==2*n-z)
# 		{printf("YES");return 0;}
# 	}
# 	printf("NO");
# }