Codeforces 697A - Pineapple Incident

A. Pineapple Incident

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

3 10 4

output

NO

input

3 10 3

output

YES

input

3 8 51

output

YES

input

3 8 52

output

YES

Note

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.

通过写出条件可以得知，所要求的Xn一定属于以下区间的时候是YES

t<=x1<=t+s<=x2<=t+s+1<=x3<=t+2s....

可以得出

0<=x-t-n*s<=s 即当(x-t)%s==0时或0<=x-t<=1是在bark time的

还有一种情况是x<t时是一定不在的，注意特判一下就好。

#include<stdio.h>#include<iostream>using namespace std;int main(){    int t,x,s;    cin>>t>>s>>x;    if (((x-t)%s==0&&x-t>=0)||x>=t&&(x-t-1)%s==0&&(x-t-1)!=0)    {        printf("YES");    }else    {        printf("NO");    }}