codeforces #362（div.2）A.Pineapple Incident【水题】

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本文介绍了一种特殊的“松果”现象，它能在特定的时间间隔内发出叫声，并探讨了如何判断这种叫声是否会在某一时刻出现，进而决定角色Barney是否能顺利吃到这个松果。

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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.

$\text{[math]}$

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.

题目大意：初始给你一个数t，然后给你一个数s，问能否在这样的序列中找到X：t，t+s，t+s+1，t+2s，t+2s+1........................‘

思路：

1、如果t==x，输出YES

2、如果t>x，输出NO

3、如果t<x,我们将X-a后对s求余，如果等于0或者等于1，输出YES。否则输出NO

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll long long int
int main()
{
    ll a,b,c;
    scanf("%I64d%I64d%I64d",&a,&b,&c);
    if(a>c)
    {
        printf("NO\n");
    }
    if(a==c)
    {
        printf("YES\n");
    }
    if(a<c)
    {
        if(((c-a)%b==0)||((c-a)%b==1)&&(c-a)!=1)
        {
            printf("YES\n");
        }
        else printf("NO\n");
    }
}