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)

#include <bits/stdc++.h>
#define rep(i,n) for(int64_t i=0;i < (int64_t)(n);i++)
#define invrep(i,n) for(int64_t i=n-1;i >= 0L;i--)
#define prep(i,n) for(int64_t i=1;i <= (int64_t)(n);i++)
using namespace std;

//////全探索したいときに開けよ////////
vector<vector<int>> fspattern;
vector<int> fspattern_draft;
int fspcnt = 0;

void setfspattern_core(int floor,int number,int degree){
  if(floor==degree){
    fspattern.at(fspcnt) = fspattern_draft;
    fspcnt++;
  }else{
    rep(i,number){
      fspattern_draft.at(floor) = i;
      setfspattern_core(floor+1,number,degree);
    }
  }
}

void setfspattern(int number,int degree){
  int64_t array_num = 1;
  fspcnt = 0;
  rep(i,degree){
    array_num *= number;
  }
  fspattern = vector<vector<int>>(array_num,vector<int>(degree));
  fspattern_draft = vector<int>(degree);
  setfspattern_core(0,number,degree);
}

//////nCrのmodを求めたいときに開けよ/////////
//int64_t univMod = 1000000007;
//int64_t factnum = 100000
//vector<int64_t> modfact(factnum);
//vector<int64_t> modinvfact(factnum);

/*void setmodfact(){
  for(int i=0;i<factnum;i++){
    if(i == 0){
      modfact.at(i) = 1;
    }else{
      modfact.at(i) = (modfact.at(i-1)*i)%univMod;
    }
  }
}

int64_t calcmodpower(int64_t a,int64_t n){
  int64_t res = 1;
  while(n != 0){
    if(n & 1){
      res = (res * a) % univMod;
    }
    a = (a*a) % univMod;
    n = n >> 1;
  }
  return res;
}

int64_t calcinverse(int64_t n){
	return calcmodpower(n,univMod-2);
}

void setmodinvfact(){
  for(int i=0;i<factnum;i++){
    if(i==0){
      modinvfact.at(i) = 1;
    }else{
      modinvfact.at(i) = calcmodpower(modfact.at(i),univMod-2);
    }
  }
}

int64_t calcmodcomb(int64_t n,int64_t r){
  return (((modfact.at(n)*modinvfact.at(r))%univMod)*modinvfact.at(n-r))%univMod;
}

int64_t calcmodpositive(int64_t n){
  return (n + ((abs(n)+univMod-1)/univMod)*univMod)%univMod;
}
  
    */


/*
int64_t calcGCD(int64_t a,int64_t b){
	int64_t p = a;
	int64_t q = b;
	int64_t c = a%b;
	while(c != 0){
		p = q;
		q = c;
		c = p%q;
	}
	return q;
}
*/

int main(){
  int a,b,x;
  cin >> a >> b >> x;
  if(a <= x && x <= a+b){
    cout << "YES" << endl;
  }else{
    cout << "NO" << endl;
  }
    
}