Problem:
Score : 100 points

Problem Statement

We have two bottles for holding water.

Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.

Bottle 2 contains C milliliters of water.

We will transfer water from Bottle 2 to Bottle 1 as much as possible.

How much amount of water will remain in Bottle 2?

Constraints

All values in input are integers.

1 \leq B \leq A \leq 20

1 \leq C \leq 20

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.

Sample Input 1

6 4 3

Sample Output 1

1

We will transfer two milliliters of water from Bottle 2 to Bottle 1, and one milliliter of water will remain in Bottle 2.

Sample Input 2

8 3 9

Sample Output 2

4

Sample Input 3

12 3 7

Sample Output 3

0

Solution in C (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef long long ll;

int main()
{
	int N;
	scanf("%d", &N);
	long h[N];
	int i = 0;
	while (i < N)
	{
		scanf("%ld", &h[i]);
		// printf("%ld ", h[i]);
		i++;
	}
	i = 0;
	while (i < N - 1)
	{
		if (h[i]- h[i+1] >= 2)
		{
			printf("No"); return 0;
		}
		if (i != N - 2 && h[i] - h[i+1] == 1)
		{
			if (i != 0 && h[i] < h[i - 1])
			{
				printf("No"); return 0;
			}
		}
		if (i == N - 2)
		{
			if (h[i] - h[i+1] == 1)
			{
				if (h[i-1] >= h[i])
				{
					printf("No"); return 0;
				}
			}
		}
		i++;
	}
	printf("Yes\n");
	return (0);
}