D. Contest Balloons
                 time limit per test
3 seconds
               memory limit per test
256 megabytes
input
standard input
output
standard output

One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.

You should know that it's important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn't considered in the standings.

A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weightwi. It's guaranteed thatti doesn't exceedwi so nobody floats initially.

Limak is a member of the first team. He doesn't like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can't give away more balloons than his team initially has.

What is the best place Limak can get?

Input

The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.

The i-th of n following lines contains two integers ti andwi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.

Output

Print one integer denoting the best place Limak can get.

Examples
Input 
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output 
3
Input 
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output 
2
Input 
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output 
2
Note

In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32,40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:

  1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
  2. Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
  3. Limak gives 1 balloon to each of two teams with 16 balloons initially.
  4. Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
  5. There are three other teams left and their numbers of balloons are 40, 14 and 2.
  6. Limak gets the third place because there are two teams with more balloons.

In the second sample, Limak has the second place and he can't improve it.

In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team6 and team 7).

题目大意：有n个队伍，每个队伍有t个气球，重量为w，当气球个数超过重量时就飞走。要将自己的气球给别人，使得自己的名次达到最大。

分析：用结构体，将其按气球数量从大到小排序，气球数量相同时按重量与气球差值大小从小到大排，再使用优先队列求出。


#include<stdio.h>
#include<algorithm>
#include<queue>
using namespace std;
struct text{
	long long num, wei;
	long long dis;
};
struct text_cmp{
	bool operator()(text a, text b)
	{
		return a.dis > b.dis;
	}
};
bool cmp(text& a,text& b)
{
	if (a.num == b.num)return a.dis<b.dis;
	else return a.num>b.num;
}
text a[300000];
priority_queue<text,vector<text>,text_cmp>q;
int main()
{
	int n,i;
	scanf("%d", &n);
	
	for (i = 0; i < n;i++)
	{
		scanf("%lld %lld", &a[i].num, &a[i].wei);
		a[i].dis = a[i].wei - a[i].num+1;
		
	}

	sort(a + 1, a + n, cmp);
	for (i = 1; i<n; i++)
	{
		if (a[i].num>a[0].num)
		{
			q.push(a[i]);
		}
		else
		{
			break;
		}
	}
	int temp=i, min=i;
	int j=i;
	while (!q.empty())
	{
		text c;
		c = q.top();
		a[0].num -= c.dis;
		q.pop();
		if (a[0].num < 0)
			break;
		temp--;
		for (; j < n; j++)
		{
			if (a[j].num>a[0].num)
			{
				q.push(a[j]);
				temp++;
			}
			else
			{
				break;
			}
			min = min<temp ? min : temp;
		}
		
	}
	//printf("%d %d\n", min, temp);
	if (temp > min)
		printf("%d\n", min);
	else
		printf("%d\n", temp);
	return 0;
}