Ted has a pineapple. This pineapple is able to bark like a bulldog! At time t (in seconds) it barks for the first time. Then every s seconds after it, it barks twice with 1 second interval. Thus it barks at times t, t + s, t + s + 1, t + 2s, t + 2s + 1, etc.

Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time x (in seconds), so he asked you to tell him if it's gonna bark at that time.

The first and only line of input contains three integers t, s and x (0 ≤ t, x ≤ 10⁹, 2 ≤ s ≤ 10⁹) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively.

Print a single "YES" (without quotes) if the pineapple will bark at time x or a single "NO" (without quotes) otherwise in the only line of output.

In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3.

In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52.

题意： t, t + s, t + s + 1, t + 2s, t + 2s + 1,   给你t，s ，x 判断x是否为序列中的值

题解： 水题也挂终测   靠hack 上分

 #include<bits/stdc++.h>

 #define ll __int64

 #define mod 1e9+7

 #define PI acos(-1.0)

 #define bug(x) printf("%%%%%%%%%%%%%",x);

 using namespace std;

 int t,s,x;

 int main()

 {

     scanf("%d %d %d",&t,&s,&x);

     if(t==x)

     {

     cout<<"YES"<<endl;

     return ;

     }

     if(x<t||x<t+s)

     {

         cout<<"NO"<<endl;

         return ;

     }

     x=x-t;

     if(x%s==||x%s==)

     {

         cout<<"YES"<<endl;

         return ;

     }

     cout<<"NO"<<endl;

     return ;

 }