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 Python (3.4.3) (judge verdict: Wrong Answer)

# -*- coding: utf-8 -*-

import sys
import os


DEBUG = False

N, L, T = map(int, input().split())
# N あり
# L 長さ
# T 時間

X = []
W = [] # list of 1 or -1
for n in range(N):
    x, w = map(int, input().split())
    X.append(x)
    if w == 1:
        W.append(1)
    else:
        W.append(-1)

if DEBUG:
    print('アリの数', N)
    print('長さ', L)
    print('時間', T)
    print('位置', X)
    print('向き', W)

# 0番目のアリがT秒後にどこにいるか
x = (X[0] + W[0] * T) % L
if x < 0:
    x += L

# T秒後のすれ違い回数
cross_num = 0
for i in range(1, N):
    # direction
    if W[0] != W[i]:
        x_diff = X[0] - X[i]
        if x_diff < 0:
            x_diff += L

        # 少なくとも1回はすれ違う
        if x_diff / 2 <= T:
            cross_num += 1

            # あと何回すれ違うか
            rest = T - x_diff / 2
            num = rest // T
            cross_num += num

            # 時刻Tでピッタリスレ違いのときはゼッケンを交換しない
            # if rest // T == rest / T:
            #     print('ピッタリ')
            #     cross_num -= 1

if DEBUG:
    print('すれ違い回数', int(cross_num))

# ゼッケン番号が衝突回数だけ増える
number = int(0 + W[0] * cross_num)
number = number % N

if DEBUG:
    print('アリ[0] 位置', x, 'ゼッケン番号', number)

# T秒後のアリの位置リスト
Y = []
if DEBUG:
    print('初期位置リスト', X)
for i, initial_x in enumerate(X):
    future_x = (initial_x + W[i] * T) % L
    Y.append(future_x)
if DEBUG:
    print('T秒後の位置リスト', Y)
Y.sort()

# 0番目のアリはソートされたYでどこ
zero_ant_index = Y.index(x)
crop = zero_ant_index - number
Z = Y[crop:] + Y[:crop]

for v in Z:
    print(v)