#!/usr/bin/env python3

# Challenge 291
#
# Task 1: Middle Index
# Submitted by: Mohammad Sajid Anwar
#
# You are given an array of integers, @ints.
#
# Write a script to find the leftmost middle index (MI) i.e. the smallest
# amongst all the possible ones.
#
#     A middle index is an index where ints[0] + ints[1] + ... + ints[MI-1]
#     == ints[MI+1] + ints[MI+2] + ... + ints[ints.length-1].
#
# If MI == 0, the left side sum is considered to be 0. Similarly,
# if MI == ints.length - 1, the right side sum is considered to be 0.
#
# Return the leftmost MI that satisfies the condition, or -1 if there is no
# such index.
# Example 1
#
# Input: @ints = (2, 3, -1, 8, 4)
# Output: 3
#
# The sum of the numbers before index 3 is: 2 + 3 + -1 = 4
# The sum of the numbers after index 3 is: 4 = 4
#
# Example 2
#
# Input: @ints = (1, -1, 4)
# Output: 2
#
# The sum of the numbers before index 2 is: 1 + -1 = 0
# The sum of the numbers after index 2 is: 0
#
# Example 3
#
# Input: @ints = (2, 5)
# Output: -1
#
# There is no valid MI.

import sys

def middle_index(n):
    for i in range(len(n)):
        s1 = sum(n[0:i])
        s2 = sum(n[i+1:])
        if s1 == s2:
            return i
    return -1

print(middle_index(list(map(int, sys.argv[1:]))))