codeforces——230A——Dragons

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我是靠谱客的博主怡然豆芽，最近开发中收集的这篇文章主要介绍codeforces——230A——Dragons，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat alln dragons that live on this level. Kirito and the dragons havestrength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equalss.

If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strengthx_i, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase byy_i.

Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.

Input

The first line contains two space-separated integers s andn (1 ≤ s ≤ 10⁴,1 ≤ n ≤ 10³). Thenn lines follow: the i-th line contains space-separated integers x_i and y_i (1 ≤ x_i ≤ 10⁴,0 ≤ y_i ≤ 10⁴) — thei-th dragon's strength and the bonus for defeating it.

Output

On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.

Examples

Input

2 2
1 99
100 0

Output

YES

Input

10 1
100 100

Output

NO

Note

In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.

In the second sample Kirito's strength is too small to defeat the only dragon and win.

第一行第一个数是Kirito的力量，第二个是n代表接下来有n组数据，每组数据由龙的力量，和战胜龙后获得的力量组成。问能不能赢所有龙

水啊水

#include<stdio.h>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
using namespace std;
struct AP
{
int com,add;
} a[1008];
bool compare(AP a,AP b);
int main()
{
int pow,n;
while(~scanf("%d%d",&pow,&n))
{
for(int i=0; i<n; i++)
cin>>a[i].com>>a[i].add;
sort(a,a+n,compare);
for(int i=0; i<n; i++)
{
if(pow==0)
continue;
if(pow>a[i].com)
pow+=a[i].add;
else
pow=0;
}
if(pow!=0)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}
bool compare(AP a,AP b)
{
return a.com<b.com;
}