""" Homework
1) Check if there is a quadruple such that:
(i, j, k, z): i < j < k < z and a[i] + a[j] = a[k] + a[z]
N <= 2000
2) Check if there is a quadruple such that:
(i, j, k, z): i < j < k < z and a[i] + a[k] = a[j] + a[z]
N <= 2000
3) Check if there is a quadruple such that:
(i, j, k, z): i < j < k < z and a[i] * a[k] = a[j] * a[z], https://community.topcoder.com/stat?c=problem_statement&pm=13951
N <= 2000
4) Find the maximum side of a cross of 1's in a matrix, N <= 2000. Think about a harder version: https://codeforces.com/problemset/problem/677/E; Don't care about the big number part.
"""

#####
# 1 #
#####

a = [int(i) for i in input().split(' ')]

def exists(a):
    n = len(a)
    d = {}
    for i in range(n):
        for j in range(i + 1, n):
            s = a[i] + a[j]
            if s not in d:
                d[s] = j

    for z in range(n - 1, -1, -1):
        for k in range(z - 1, -1, -1):
            s = a[z] + a[k]
            if s in d and d[s] < k:
                return True
    return False

print(exists(a))

#####
# 2 #
#####

a = [int(i) for i in input().split(' ')]

def exists(a):
    n = len(a)
    d = {}
    for i in range(n):
        for j in range(i + 1, n):
            s = a[i] - a[j]
            if s not in d:
                d[s] = j

    for z in range(n - 1, -1, -1):
        for k in range(z - 1, -1, -1):
            s = a[z] - a[k]
            if s in d and d[s] < k:
                return True
    return False

print(exists(a))