Problem:
Score: 400 points

Problem Statement

Kizahashi, who was appointed as the administrator of ABC at National Problem Workshop in the Kingdom of AtCoder, got too excited and took on too many jobs.

Let the current time be time 0. Kizahashi has N jobs numbered 1 to N.

It takes A_i units of time for Kizahashi to complete Job i. The deadline for Job i is time B_i, and he must complete the job before or at this time.

Kizahashi cannot work on two or more jobs simultaneously, but when he completes a job, he can start working on another immediately.

Can Kizahashi complete all the jobs in time? If he can, print Yes; if he cannot, print No.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

1 \leq A_i, B_i \leq 10^9 (1 \leq i \leq N)

Input

Input is given from Standard Input in the following format:

N
A_1 B_1
.
.
.
A_N B_N

Output

If Kizahashi can complete all the jobs in time, print Yes; if he cannot, print No.

Sample Input 1

5
2 4
1 9
1 8
4 9
3 12

Sample Output 1

Yes

He can complete all the jobs in time by, for example, doing them in the following order:

Do Job 2 from time 0 to 1.

Do Job 1 from time 1 to 3.

Do Job 4 from time 3 to 7.

Do Job 3 from time 7 to 8.

Do Job 5 from time 8 to 11.

Note that it is fine to complete Job 3 exactly at the deadline, time 8.

Sample Input 2

3
334 1000
334 1000
334 1000

Sample Output 2

No

He cannot complete all the jobs in time, no matter what order he does them in.

Sample Input 3

30
384 8895
1725 9791
170 1024
4 11105
2 6
578 1815
702 3352
143 5141
1420 6980
24 1602
849 999
76 7586
85 5570
444 4991
719 11090
470 10708
1137 4547
455 9003
110 9901
15 8578
368 3692
104 1286
3 4
366 12143
7 6649
610 2374
152 7324
4 7042
292 11386
334 5720

Sample Output 3

Yes

Solution in Python (3.4.3) (judge verdict: Accepted)

import math
import fractions
import bisect
import collections
import itertools
import heapq
import string
import sys
sys.setrecursionlimit(10**9)
def gcd(a,b):return fractions.gcd(a,b) #最大公約数
def lcm(a,b):return (a*b) // fractions.gcd(a,b) #最小公倍数
def iin(): return int(input()) #整数読み込み
def isn(): return input().split() #文字列読み込み
def imn(): return map(int, input().split()) #整数map取得
def iln(): return list(map(int, input().split())) #整数リスト取得
def iln_s(): return sorted(iln()) # 昇順の整数リスト取得
def iln_r(): return sorted(iln(), reverse=True) # 降順の整数リスト取得
def join(l, s=''): return s.join(l) #リストを文字列に変換
def perm(l, n): return itertools.permutations(l, n) # 順列取得
def perm_count(n, r): return math.factorial(n) // math.factorial(n-r) # 順列の総数
def comb(l, n): return itertools.combinations(l, n) # 組み合わせ取得
def comb_count(n, r): return math.factorial(n) // (math.factorial(n-r) * math.factorial(r)) #組み合わせの総数
def divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i: divisors.append(n//i)
    return divisors
def is_prime(n):
    if n == 1: return False
    for i in range(2, int(n**0.5) + 1):
        if n % i == 0: return False 
    return True
def primes(n):
    is_prime = [True] * (n+1)
    is_prime[0] = False
    is_prime[1] = False
    for i in range(2, int(n**0.5)+1):
        if not is_prime[i]: continue
        for j in range(i * 2, n + 1, i): is_prime[j] = False
    return [i for i in range(n+1) if is_prime[i]]
def sieve_of_e(n):
    is_prime = [True] * (n+1)
    is_prime[0] = False
    is_prime[1] = False
    for i in range(2, int(n**0.5)+1):
        if not is_prime[i]: continue
        for j in range(i * 2, n+1, i): is_prime[j] = False
    return is_prime
def run_length_encode(s):
    count = 1
    code = []
    target = s[0]
    for i in range(1, len(s)):
        if target == s[i]: count += 1
        else:
            code.append((target, count))
            target = s[i]
            count = 1
    code.append((target, count))
    return code

N = iin()
AB = sorted([iln() for _ in range(N)], key=lambda x:x[1])

time = 0
for i in range(N):
    time += AB[i][0]
    if AB[i][1] < time:
        print('No')
        break
else:
    print('Yes')