Visiting a Friend CodeForces - 902A

最新推荐文章于 2021-07-18 10:31:46 发布

原创最新推荐文章于 2021-07-18 10:31:46 发布 · 243 阅读

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本文介绍了一个有趣的编程问题：猪能否仅通过使用特定位置的传送器从家到达朋友家。通过分析传送器的位置和限制，文章提供了一种算法来确定是否可以实现这一目标。

Pig is visiting a friend.

Pig's house is located at point 0, and his friend's house is located at point m on an axis.

Pig can use teleports to move along the axis.

To use a teleport, Pig should come to a certain point (where the teleport is located) and choose where to move: for each teleport there is the rightmost point it can move Pig to, this point is known as the limit of the teleport.

Formally, a teleport located at point x with limit y can move Pig from point x to any point within the segment [x; y], including the bounds.

$\text{[math]}$

Determine if Pig can visit the friend using teleports only, or he should use his car.

Input

The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of teleports and the location of the friend's house.

The next n lines contain information about teleports.

The i-th of these lines contains two integers a_i and b_i (0 ≤ a_i ≤ b_i ≤ m), where a_iis the location of the i-th teleport, and b_i is its limit.

It is guaranteed that a_i ≥ a_i - 1 for every i (2 ≤ i ≤ n).

Output

Print "YES" if there is a path from Pig's house to his friend's house that uses only teleports, and "NO" otherwise.

You can print each letter in arbitrary case (upper or lower).

Example

Input

Output

YES

Input

Output

NO

#include <stdio.h>
#include <string.h>
#define MAX 100

int main()
{
	int n,i,j,m,x,y,max;
	while(scanf("%d%d",&n,&m) != EOF)
    {
        max = 0;
        for(i = 0;i < n;i++)
        {
            scanf("%d%d",&x,&y);
            if(x <= max)
            {
                if(y >= max)
                    max = y;
            }
        }
        if(max >= m)
            printf("YES\n");
        else
            printf("NO\n");
    }
	return 0;
}