Problem:
Score : 700 points

Problem Statement

There is a circle with a circumference of L.
Each point on the circumference has a coordinate value, which represents the arc length from a certain reference point clockwise to the point.
On this circumference, there are N ants.
These ants are numbered 1 through N in order of increasing coordinate, and ant i is at coordinate X_i.

The N ants have just started walking.
For each ant i, you are given the initial direction W_i. Ant i is initially walking clockwise if W_i is 1; counterclockwise if W_i is 2.
Every ant walks at a constant speed of 1 per second.
Sometimes, two ants bump into each other.
Each of these two ants will then turn around and start walking in the opposite direction.

For each ant, find its position after T seconds.

Constraints

All input values are integers.

1 \leq N \leq 10^5

1 \leq L \leq 10^9

1 \leq T \leq 10^9

0 \leq X_1 < X_2 < ... < X_N \leq L - 1

1 \leq W_i \leq 2

Input

The input is given from Standard Input in the following format:

N L T
X_1 W_1
X_2 W_2
:
X_N W_N

Output

Print N lines.
The i-th line should contain the coordinate of ant i after T seconds. Here, each coordinate must be between 0 and L-1, inclusive.

Sample Input 1

3 8 3
0 1
3 2
6 1

Sample Output 1

1
3
0

1.5 seconds after the ants start walking, ant 1 and 2 bump into each other at coordinate 1.5.
1 second after that, ant 1 and 3 bump into each other at coordinate 0.5.
0.5 seconds after that, that is, 3 seconds after the ants start walking, ants 1, 2 and 3 are at coordinates 1, 3 and 0, respectively.

Sample Input 2

4 20 9
7 2
9 1
12 1
18 1

Sample Output 2

7
18
18
1

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
typedef struct tuple{ll x,w,id;}E;
int psort(const void*a,const void*b){
	E*p=(E*)a,*q=(E*)b;
	if((*p).x<(*q).x)return -1;
	if((*p).x>(*q).x)return  1;
	//以下を処理するのは「ちょうどL秒経過した瞬間にすれ違う2匹」を比較するとき
	//すれ違い回数のカウントでは、ちょうどのときを回数にカウントしていないので
	//時計回りの方が-ε、反時計周りの方が+εにいると考えて良い
	if((*p).w<(*q).w)return -1;
	return 1;
}

ll n,L,T;
ll x[100010],w[100010];
E ans[100010];
int main(){
	scanf("%lld%lld%lld",&n,&L,&T);
	rep(i,0,n)scanf("%lld%lld",x+i,w+i);
	//L秒後に蟻の位置のなす集合
	rep(i,0,n){
		if(w[i]==1){
			ans[i].x=(x[i]+T)%L;
		}else{
			ans[i].x=((x[i]-T)%L+L)%L;
		}
		ans[i].w=w[i];
		ans[i].id=i;
	}
	qsort(ans,n,sizeof(E),psort);
	
	ll cnt=0;
	rep(i,1,n)if(w[0]!=w[i]){
		//大体L/2秒ごとにすれ違う
		cnt+=T/L*2;
		//残りのT%L秒で高々1回すれ違う
		if(w[0]==1){
			if(x[i]-x[0]<T%L*2)cnt++;
		}else{
			if(L-(x[i]-x[0])<T%L*2)cnt++;
		}
		cnt%=n;
	}
	
	//cnt番目の蟻は、ans[i].id==0となるiのans[i].xにいる
	ll offset=0;
	rep(i,0,n)if(ans[i].id==0){
		offset=i;
		break;
	}
	
	rep(i,0,n)printf("%lld\n",ans[(i+offset-cnt+n)%n].x);
}