codeforces-230A-Dragons（贪心+排序）

题目链接：codeforces.com/problemset/problem/230/A

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 x_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 by y_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 and n (1 ≤ s ≤ 10⁴, 1 ≤ n ≤ 10³). Then n lines follow: the i-th line contains space-separated integers x_i and y_i (1 ≤ x_i ≤ 10⁴, 0 ≤ y_i ≤ 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.

解析：

一只龙有 S 的血量。

其他 n 只龙，第 i 只龙有 Xi 的血量，其被打死之后，胜利者有 Yi 的血量奖励。

若要该龙胜利到最后，则在每一场战斗后，该龙的最优选择就是选择一只血量最小的龙战斗。

因此只要按照每只龙的血量排序即可。

参考代码：

#include
#include
struct node
{
    int x;
    int y;       
}ans[1010];
int cmp(const void *a, const void *b)
{
    struct node *aa = (node *)a;
    struct node *bb = (node *)b;
    return (((aa->x)>(bb->x))?1:-1);    
}
int main()
{
    int s, n;
    while(scanf("%d %d", &s, &n) != EOF)
    {
        for(int i=0; i