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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

//#pragma once
#include <stdio.h>
#include <string>
#include <iostream>
#include <queue>
#include <algorithm>
#include <sstream>
#include <vector>
#include <math.h>
#include <set>
#include <map>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cctype>


using namespace std;
#define modd 1000000007


typedef pair<long long, long long> pl;


typedef string::const_iterator State;
class ParseError {};




#define ll long long




/*
class UnionFind {
public:
	vector <ll> par;
	vector <ll> siz;
	vector <ll> gr;

	// Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL) ,gr(sz_){
		for (ll i = 0; i < sz_; ++i) par[i] = i;
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.assign(sz_, 1LL);  // resize だとなぜか初期化されなかった
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}

	// Member Function
	// Find
	ll root(ll x) { // 根の検索
		while (par[x] != x) {
			x = par[x] = par[par[x]]; // x の親の親を x の親とする
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;

		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		return true;
	}



	bool issame(ll x, ll y) { // 連結判定
		return root(x) == root(y);
	}

	ll size(ll x) { // 素集合のサイズ
		return siz[root(x)];
	}


};*/


long long N, M, K, a, b, c, d, e, H, W, L;
ll A[2][200005] = {};
ll B[100005] = {};
ll C[300005] = {};
bool f, ff;
string S,T;
set <long long>sll;
pl bufpl;
vector <long long>vl;
vector <pl>vpl;
vector <string> vs;
set<ll> llset;
multiset<ll> llmset;

/*struct ST
{
	long long Kati;
	int bangou;

	bool operator<(const ST& another) const
	{
		return Kati < another.Kati;//比較
	};

};



vector <ST> vst;
ST st[200005];
ST bufst;
bitset<5000> bits;*/


long long modinv(long long aa, long long mm) {
	long long bb = mm, uu = 1, vv = 0;
	while (bb) {
		long long tt = aa / bb;
		aa -= tt * bb; swap(aa, bb);
		uu -= tt * vv; swap(uu, vv);
	}
	uu %= mm;
	if (uu < 0) uu += mm;
	return uu;
}



ll minn, maxx;
double ansa;
bool fl[30] = {};


string ss, sss;


int main() {


	cin >> N;

	for (int i = 0; i < N; i++) {
		cin >> a >> b;
		bufpl.first = b;
		bufpl.second = a;
		vpl.push_back(bufpl);

	}

	sort(vpl.begin(), vpl.end());

	a = 0;
	f = true;

	for (int i = 0; i < N; i++) {
		a += vpl[i].second;
		if (a > vpl[i].first) {
			f = false;
			break;
		}

	}

	if (f) {
		cout <<"Yes"<< endl;
	}
	else {
		cout << "No" << endl;
	}
	return 0;

}