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:
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
# import numpy as np
starttime = time.time()
# import numpy as np
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]

try:
    # sys.setrecursionlimit(int(pow(10,6)))
    sys.stdin = open("input.txt", "r")
    # sys.stdout = open("../output.txt", "w")
except:
    pass

@lru_cache(None)
def rec(i,count):
    # print(i,count)
    if i==d:
        if count==st:
            global pos
            pos=1
            return True
        return False
    if count>st or count+8*(d-i)<st:
        return False
    for j in range(A[i][0],A[i][1]+1):
        if rec(i+1,count+j):
            ans.append(j)
            return True
    return False

d,st=L()
A=[L() for i in range(d)]
global pos
pos=0
ans=[]
rec(0,0)

if pos==1:
    print("YES")
    print(*ans[::-1])
else:
    print("NO")






















endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")