Problem: Tomorrow Peter has a Biology exam. He does not like this subject much, but d days ago he learnt that he would have to take this exam. Peter's strict parents made him prepare for the exam immediately, for this purpose he has to study not less than minTimei and not more than maxTimei hours per each i-th day. Moreover, they warned Peter that a day before the exam they would check how he has followed their instructions.

So, today is the day when Peter's parents ask him to show the timetable of his preparatory studies. But the boy has counted only the sum of hours sumTime spent him on preparation, and now he wants to know if he can show his parents a timetable sсhedule with d numbers, where each number sсhedulei stands for the time in hours spent by Peter each i-th day on biology studies, and satisfying the limitations imposed by his parents, and at the same time the sum total of all schedulei should equal to sumTime.

Input

The first input line contains two integer numbers d, sumTime (1 ≤ d ≤ 30, 0 ≤ sumTime ≤ 240) — the amount of days, during which Peter studied, and the total amount of hours, spent on preparation. Each of the following d lines contains two integer numbers minTimei, maxTimei (0 ≤ minTimei ≤ maxTimei ≤ 8), separated by a space — minimum and maximum amount of hours that Peter could spent in the i-th day.

Output

In the first line print YES, and in the second line print d numbers (separated by a space), each of the numbers — amount of hours, spent by Peter on preparation in the corresponding day, if he followed his parents' instructions; or print NO in the unique line. If there are many solutions, print any of them.

Examples

Input

1 48
5 7


Output

NO


Input

2 5
0 1
3 5


Output

YES
1 4 

Solution (in PYTHON3):

Solution:
# -*- coding: utf-8 -*-
"""
Created on Mon Mar  2 18:08:51 2020

@author: akprasad

4B. Before an Exam
time limit per test: 0.5 second
memory limit per test: 64 megabytes
input: standard input
output: standard output

Tomorrow Peter has a Biology exam.
He does not like this subject much,
but d days ago he learnt that he would have to take this exam.
Peter's strict parents made him prepare for the exam immediately,
for this purpose he has to study not less than minTimei
and not more than maxTimei hours per each i-th day.
Moreover, they warned Peter that a day before the exam
they would check how he has followed their instructions.

So, today is the day when Peter's parents ask him to show
the timetable of his preparatory studies. But the boy has
counted only the sum of hours sumTime spent him on preparation,
and now he wants to know if he can show his parents a timetable
sсhedule with d numbers, where each number sсhedulei stands for
the time in hours spent by Peter each i-th day on biology studies,
and satisfying the limitations imposed by his parents, and at the
same time the sum total of all schedulei should equal to sumTime.

Input
The first input line contains two integer numbers d, 
sumTime (1 ≤ d ≤ 30, 0 ≤ sumTime ≤ 240) — the amount of days,
during which Peter studied, and the total amount of hours,
spent on preparation. Each of the following d lines contains
two integer numbers minTimei, maxTimei (0 ≤ minTimei ≤ maxTimei ≤ 8),
separated by a space — minimum and maximum amount of hours that
Peter could spent in the i-th day.

Output
In the first line print YES, and in the second line print d numbers
(separated by a space), each of the numbers — amount of hours,
spent by Peter on preparation in the corresponding day,
if he followed his parents' instructions; or print NO in
the unique line. If there are many solutions, print any of them.

Examples

input
1 48
5 7
output
NO

input
2 5
0 1
3 5
output
YES
1 4 

"""

def print_solution(d, sum_time, min_h, max_h):
    # first check if possible or not
    if sum_time > sum(max_h) or sum_time < sum(min_h):
        print("NO")
        return

    schedule = min_h[:]
    extra_time = sum_time - sum(min_h)
    for i in range(d):
        if max_h[i] - min_h[i] > extra_time:  # terminal case where we don't have enough extra time left
            schedule[i] += extra_time
            break
        # normal case
        schedule[i] = max_h[i]
        extra_time -= max_h[i] - min_h[i]
    print("YES")
    print(*schedule, sep=" ")


if __name__ == '__main__':
    d, sum_time = [int(x) for x in input().split()]
    min_h, max_h = [], []
    for _ in range(d):
        min_d, max_d = [int(x) for x in input().split()]
        min_h.append(min_d)
        max_h.append(max_d)
    print_solution(d, sum_time, min_h, max_h)