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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
//#include <iomanip>
#define hell                         1000000007
#define M                            1000000007
#define Maxi                      10000000000000000
#define lcm(a,b)                     (a*b)/__gcd(a,b)
#define ll                           long long
 
#define vll                          vector<ll>
#define vi                           vector<long long>
#define pll                          vector< pair<ll,ll> >
#define pb                           push_back
//#define mp                           make_pair
#define all(v)                       v.begin(),v.end()
#define lbnd                         lower_bound
#define ubnd                         upper_bound
#define bs                           binary_search
#define F                            first  
#define S                            second
#define rep(i,a,b)                   for(i=a;i<b;i++)
#define parr(a,n)                    for(i=0;i<n;i++) cout<<a[i]<<" ";cout<<endl;
#define pcont(a)                     for(auto i:a) cout<<i<<" ";cout<<endl;
#define ret(x)                       return cout<<x,0;
#define dbg(x)                       cout << #x << " is " << x << endl;
#define endl                           '\n'
 //const ll infl=0x3f3f3f3f3f3f3f3fLL;

 #define pi                           3.141592653589793238
// Debug //
 
#define trace(x)                 cerr << #x << ": " << x << endl;
#define trace1(x)                cerr << #x << ": " << x << endl;
#define trace2(x, y)             cerr << #x << ": " << x << " | " << #y << ": " << y << endl;
#define trace3(x, y, z)          cerr << #x << ": " << x << " | " << #y << ": " << y << " | " << #z << ": " << z << endl;
#define trace4(a, b, c, d)       cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << endl;
#define trace5(a, b, c, d, e)    cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << endl;
#define trace6(a, b, c, d, e, f) cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << " | " << #f << ": " << f << endl;
using namespace std;

/*ll prime[1000006];
void fun()
{
  for(ll i=2;i<=1000000;i++)
  {
    if(prime[i]==0)
    {
      for(ll j=i;j<=1000000;j+=i)
      {
        if(prime[j]==0)
        prime[j]=i;
      }
    }
  }
}*/
/*const int MAXR = 1000005;

ll cubic_root(ll x)
{
    ll l = 0, r = MAXR;
    while (l != r) {
        ll m = (l + r + 1) / 2;
        if (m * m * m > x)
            r = m - 1;
        else
            l = m;
    }
    return l;
}*/

/*ll power(ll x,ll y,ll m) 
{ 
    if (y == 0) 
        return 1; 
    ll p = power(x, y/2, m) % m; 
    p = (p * p) % m; 
 
    return (y%2 == 0)? p : (x * p) % m; 
}*/
/*bool checkprime(ll n)
{ 
  ll i;
  if(n<=3)
    return true;

  if(n%2==0 || n%3==0)
    return false;

  for(i=5;i*i<=n;i+=6)
  {
    if(n%i==0 || n%(i+2)==0)
      return false;
  }

  return true;

}

 ll check(ll n,ll b) 
{
  ll i,p=1,P=1,ans=1;
  if(n<3)
    return 0;
  for(i=1;i<=b;i++)
  {
    ans=ans*(n-i+1/i)%hell;
  }

  return ans;
}*/

/*const int N = 1e6+1;

ll prime[N];
//ll n;
vector<ll>p;
vector<int>ans;


void sv()
{
  for(int i=1;i<N;i++)
    prime[i]=1;

  for(int i=2;i*i<=N;i++)
  {
    if(prime[i])
    {
      for(int j=i*i;j<N;j+=i)
      {
        prime[j]=0;
      }
    }
  }

  for(int i=2;i<N;i++)
  {
    if(prime[i])
      p.pb(i);
  }
  return ;
}
*/
/*void solve()
{
  int n;
  cin>>n;
  sv();
  for(auto i:p)
  {
    if(i>n)
      break;

    for(int j=1;pow(i,j)<=n;j++)
      ans.pb(pow(i,j));
  }

  cout<<ans.size()<<endl;
  for(auto i:ans)
    cout<<i<<" ";
  cout<<endl;
  //cout<<"ef"<<endl;
return ;
}*/
/*ll m;
long long fast_power(long long base, long long power) {
    long long result = 1;
    while(power > 0) {

        if(power % 2 == 1) { // Can also use (power & 1) to make code even faster
            result = (result*base) % m;
        }
        base = (base * base) % m;
        power = power / 2; // Can also use power >>= 1; to make code even faster
    }
    return result;
}*/
 

int main()
{
      ios_base::sync_with_stdio(false);
      cin.tie(NULL);cout.tie(NULL);
      ll n;
      cin>>n;
      vector<pair<ll,ll>>v;
      ll i;
      for(i=0;i<n;i++)
      {
        ll a,b;
        cin>>a>>b;
        v.pb({b,a});
      }
      bool ok=true;
      sort(all(v));
      for(i=1;i<n;i++)
      {
        v[i].S+=v[i-1].S;
      }
      for(i=0;i<n;i++)
      {
        if(v[i].S>v[i].F)
        {
            ok=false;
            break;
        }
      }
      if(ok)
        cout<<"Yes"<<endl;
      else
        cout<<"No"<<endl;
}