Pineapple Incident

Description

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.

Input

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.

Output

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.

Sample Input

Input

3 10 4

Output

NO

Input

3 10 3

Output

YES

Input

3 8 51

Output

YES

Input

3 8 52

Output

YES

Hint

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.

很简单的题,只要做个判断(x-t)%s=1||0就ok了,还有2个特别判断,就是当x=t+1和x<t时的情况.

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){long long t,s,x,i;while(scanf("%lld%lld%lld",&t,&s,&x)!=EOF){int flag=0;if(x==t)flag=1;else if(x<t)flag=0;else if(x==t+1)flag=0;else {if((x-t)%s==1||(x-t)%s==0)flag=1;}if(flag==0)printf("NO\n");elseprintf("YES\n");}return 0;}