public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        
        int[] array = new int[n];
        
        for(int i = 0; i < n; i++) {
            array[i] = sc.nextInt();
        }
        
        System.out.println(countSubarraysWithNegativeSums(array));
        
        sc.close();
    }
    
    private static int countSubarraysWithNegativeSums(int[] array) {
        int count = 0;
        int currentSum;
        
        // Look all the possible subArrays and check it sums
        for(int i = 1; i <= array.length; i++) {
            for(int j = 0; j <= array.length - i; j++) {
                currentSum = 0;
                for(int z = 0; z < i; z++) {
                    currentSum += array[j + z];
                }
                
                if(currentSum < 0)
                    count++;
            }
        } 
    
    return count;
}
				
				
//Second Approach: Time Complexity O(n^2)

private static int countSubarraysWithNegativeSums(int[] array) {
    int count = 0;
    int currentSum;
    int n = array.length;
    
    // Look all the possible subArrays and check 
    for (int start = 0; start < n; start++) {
        currentSum = 0;
        for (int end = start; end < n; end++) {
            currentSum += array[end];
            
            if (currentSum < 0)
                count++;
            
        }
    }
		
        return count;
    }