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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repe(i, n) for (int i = 0; i <= (int)(n); i++)
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl
#define ANS cout << ans << endl
#define all(a) a.begin(),a.end()
#define SORT(a) sort(all(a))
#define REVERSE(a) reverse(all(a))
#define int long long
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORe(i, m, n) for(int i = m;i <= n;i++)
#define VIin(a) for(int i=0; i<a.size(); i++) cin >> a[i]
#define OUT(a) cout << a << endl
typedef vector<int> VI;
typedef vector<VI> VVI;
const int MOD=1e9+7;
const long long INF = 1LL << 60;
/*最大公約数*/int gcd(int x, int y) { return y ? gcd(y,x%y) : x;}
/*最小公倍数*/int lcm(int a, int b) { int g = gcd(a, b); return a / g * b; }
/*素数判定*/int isPrime(int n){if(n<2){return 0;}for(int i=2; i*i<=n; i++){if(n%i==0){return 0;}}return 1;}
/*約数列挙*/vector<int> getDivisor(int n){vector<int> v; for(int i=1; i*i<=n; i++){if(n%i==0){v.push_back(i);if(i!=n/i){v.push_back(n/i);}}}sort(v.begin(),v.end()); return v;}
//----------------------------------------------
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    
    int a,b,x; cin >> a >> b >> x;
    if(x-a < b && x > a) YES;
    else NO;

    return 0;
}