Codeforces Round #353 (Div. 2) A. Infinite Sequence Thinking problem

A. Infinite sequencetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Vasya likes everything infinite. Now he is studying the properties of a sequences, such that it first element is equal toa( s1? =? a ), and the difference between any and neighbouring elements are equal toC( si?-? s i?-? 1? =? C ). In particular, Vasya wonders if his favourite integerbAppears in this sequence, which is, there exists a positive integerI, such that si? =? b . Of course, you is the person who he asks for a help.

Input

The first line of the input contain three integersa,bandC(?-?9? ≤? a,? b,? c. ≤?109 )-the first element of the sequence, Vasya ' favorite number and the difference between any of neighbouring elements of The sequence, respectively.

Output

If b appears in the sequence s print "YES" (without quotes), otherwise print "NO" (Wit Hout quotes).

Examplesinput

1 7 3

Output

YES

Input

10 10 0

Output

YES

Input

1-4 5

Output

NO

Input

0 60 50

Output

NO

Note

In the first sample, the sequence starts from integers 1, 4, 7, and so 7 are its element.

In the second sample, the favorite an integer of Vasya is equal to the first element of the sequence.

In the third sample all elements of the sequence is greater than Vasya ' s favorite integer.

In the fourth sample, the sequence starts from 0, +, and all the following elements is G Reater than Vasya ' s favorite integer.

Source

A. Infinite Sequence

My Solution

by Cha, (┬_┬), the reason is to use (b-a) * C >= (b-a) with the same number of C, or b-a = = 0. here int * int overflow, later still basically do not, write.

Honest ((t>=0&&c > 0) | | (t<=0 && C < 0)) Write it like this.

If c = = 0, then a = = B Yes otherwise NO

If c! = 0, then see if it can be divisible, and (b-a) the same number as C (if B is not equal to a)

#include <iostream> #include <cstdio>using namespace Std;int main () {    int A, b, C;    scanf ("%d%d%d", &a, &b, &c);    int t = b-a;    if (c = = 0) {        if (a = = b) printf ("YES");        else printf ("NO");    }    else{        //if (t%c = = 0 && (t*c >= 0)) printf ("YES"),//!t*c overflow        if (t%c = = 0 && ((t>=0&& C > 0) | | (t<=0 && C < 0))) printf ("YES");        else printf ("NO");    }    return 0;}

Thank you!

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Codeforces Round #353 (Div. 2) A. Infinite Sequence Thinking problem