Problem:
Score : 100 points

Problem Statement

There are a total of A + B cats and dogs.
Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs.

Determine if it is possible that there are exactly X cats among these A + B animals.

Constraints

1 \leq A \leq 100

1 \leq B \leq 100

1 \leq X \leq 200

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B X

Output

If it is possible that there are exactly X cats, print YES; if it is impossible, print NO.

Sample Input 1

3 5 4

Sample Output 1

YES

If there are one cat and four dogs among the B = 5 animals, there are X = 4 cats in total.

Sample Input 2

2 2 6

Sample Output 2

NO

Even if all of the B = 2 animals are cats, there are less than X = 6 cats in total.

Sample Input 3

5 3 2

Sample Output 3

NO

Even if all of the B = 3 animals are dogs, there are more than X = 2 cats in total.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#pragma comment(linker, "/STACK:16777216")

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <iostream>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <numeric>
#include <algorithm>
#include <utility>
#include <bitset>
#include <cmath>
#include <sstream>
#include <functional>
#include <iomanip>

#define all(a) (a).begin(),(a).end()
#define sz(a) (int)(a).size()
#define eps (1e-15)
using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector< vi > vvi;
typedef vector<double> vd;
typedef vector< vd > vvd;
typedef vector< string > vs;
typedef pair< int, int > pii;
typedef vector< pii > vpii;

vector<long long> p_pow;
const int p = 61;

long long myhash(string s)
{
	long long res = 0;
	for (int i = 0; i < sz(s); i++)
	{
		res += (s[i] - 'a' + 1) * p_pow[i];

	}
	return res;
}

int main()
{
#ifdef LOCAL_MACHINE
	freopen("input.txt", "rt", stdin);
	freopen("output.txt", "wt", stdout);
#endif

	int a, b, x;
	cin >> a >> b >> x;
	if (x < a || (a + b) < x)
		cout << "NO";
	else
		cout << "YES";
	return 0;

}