Turn the Rectangles 1008B

output

standard output

There are $\mathrm{nn rectangles\; in\; a\; row.\; You\; can\; either\; turn\; each\; rectangle\; by 9090 degrees\; or\; leave\; it\; as\; it\; is.\; If\; you\; turn\; a\; rectangle,\; its\; width\; will\; be\; height,\; and\; its\; height\; will\; be\; width.\; Notice\; that\; you\; can\; turn\; any\; number\; of\; rectangles,\; you\; also\; can\; turn\; all\; or\; none\; of\; them. You\; can\; not\; change\; the\; order\; of\; the\; rectangles.}$

Find out if there is a way to make the rectangles go in order of non-ascending height. In other words, after all the turns, a height of every rectangle has to be not greater than the height of the previous rectangle (if it is such).

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1051\le n\le 105) —\; the\; number\; of\; rectangles.}$

Each of the next $\mathrm{nn lines\; contains\; two\; integers wiwi and hihi (1\le wi,hi\le 1091\le wi,hi\le 109) —\; the\; width\; and\; the\; height\; of\; the ii-th\; rectangle.}$

Output

Print "YES" (without quotes) if there is a way to make the rectangles go in order of non-ascending height, otherwise print "NO".

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

Examples

input

Copy

3
3 4
4 6
3 5

output

Copy

YES

input

Copy

2
3 4
5 5

output

Copy

NO

Note

In the first test, you can rotate the second and the third rectangles so that the heights will be [4, 4, 3].

In the second test, there is no way the second rectangle will be not higher than the first one.

题意：这里有n个矩形，你可以让每个矩形旋转90度，可以旋转多次或者不旋转，但是你不能改变矩形的排列顺序，每一个矩形不能比它前一个矩形高，如果可以输出YES，否则输出NO。

分析：使第一个矩形的高最大，如果下一个矩形的宽和高的最小值小于它前一个矩形的高，比较：1，(1)如果宽大于前一个的高，则把该矩形的高作为高(2)如果高大于前一个矩形的高，则把该矩形的宽作为该矩形的高(3)如果宽和高均小于前一个的高，则找出宽和高较大的那个作为该矩形的高   2，如果该矩形的宽高最小值大于前一个矩形的高，那么直接输出NO

#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
    int n,w[100005],h[100005];
    while(~scanf("%d",&n))
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d %d",&w[i],&h[i]);
        }
        int t1,t2,i;
        for(i=1; i<=n; i++)
        {
            if(i==1)
                t1=max(w[i],h[i]);
            else
            {
                t2=min(w[i],h[i]);
                if(t2<=t1)
                {
                    if(w[i]>t1)
                        t1=h[i];
                    else if(h[i]>t1)
                        t1=w[i];
                    else if(w[i]<=t1&&h[i]<=t1)
                        t1=max(w[i],h[i]);
                }
                else
                {
                    break;
                }
            }

        }
        if(i==n+1)
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}