Dragons

http://codeforces.com/problemset/problem/230/A

Dragons

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, 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 equals s.

If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, 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 by yi.

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 and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-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.

Sample test(s)

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 to 2 + 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.

题意就是一个人杀龙，这个人初始力量为s，有n条龙，可以按任何顺序杀龙，杀死一条龙有力量奖励y，龙的力量为x，当这个人的力量s大于龙的力量x时才能杀死这条龙，然后他的力量增加这条龙的奖励y，即s+=y

所以直接贪心就行了。

AC代码：

#include<iostream>
#include<cstdio>
#include<algorithm>

using namespace std;

struct Node
{
    int x,y;
}g[1010];

bool cmp(Node a, Node b)
{
    return a.x < b.x;
}

int main()
{
    int n,s,i;
    while(scanf("%d%d",&s,&n)!=EOF)
    {
        for(i = 0; i < n; i++)
        {
            scanf("%d%d",&g[i].x,&g[i].y);
        }
        sort(g,g+n,cmp);
        for(i = 0; i < n; i++)
        {
            if(s > g[i].x)
            {
                s+=g[i].y;
            }
            else
            {
                break;
            }
        }
        if(i == n)
        {
            printf("YES
");
        }
        else
        {
            printf("NO
");
        }
    }

    return 0;
}