During the lesson small girl Alyona works with one famous spreadsheet computer program and learns how to edit tables.

Now she has a table filled with integers. The table consists of n rows and m columns. By ai, j we will denote the integer located at the i-th row and the j-th column. We say that the table is sorted in non-decreasing order in the column j if ai, j ≤ ai + 1, j for all i from 1 to n - 1.

Teacher gave Alyona k tasks. For each of the tasks two integers l and r are given and Alyona has to answer the following question: if one keeps the rows from l to r inclusive and deletes all others, will the table be sorted in non-decreasing order in at least one column? Formally, does there exist such j that ai, j ≤ ai + 1, j for all i from l to r - 1 inclusive.

Alyona is too small to deal with this task and asks you to help!

Input
The first line of the input contains two positive integers n and m (1 ≤ n·m ≤ 100 000) — the number of rows and the number of columns in the table respectively. Note that your are given a constraint that bound the product of these two integers, i.e. the number of elements in the table.

Each of the following n lines contains m integers. The j-th integers in the i of these lines stands for ai, j (1 ≤ ai, j ≤ 109).

The next line of the input contains an integer k (1 ≤ k ≤ 100 000) — the number of task that teacher gave to Alyona.

The i-th of the next k lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n).

Output
Print “Yes” to the i-th line of the output if the table consisting of rows from li to ri inclusive is sorted in non-decreasing order in at least one column. Otherwise, print “No”.

Example
Input
5 4
1 2 3 5
3 1 3 2
4 5 2 3
5 5 3 2
4 4 3 4
6
1 1
2 5
4 5
3 5
1 3
1 5
Output
Yes
No
Yes
Yes
Yes
No
Note
In the sample, the whole table is not sorted in any column. However, rows 1–3 are sorted in column 1, while rows 4–5 are sorted in column 3.
心得：
题目大致意思就是输入一个矩阵 k次询问 每次询问l和r行之间是否存在非降序数列。
这里要用到vector 来记录大数量的二维数组。
用p和f数组记录升序数的位置。
特别巧妙的做法。
如果用d二维数组记录，那就会T。

#include<stdio.h>
#include<vector>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=100005;
vector<int>a[maxn];
int p[maxn];
int f[maxn];
int main()
{
   
	int m,n;
	a[0].resize(maxn,0);
	scanf("%d%d",&m,&n);
	for(int i=1;i<=m;i++)
	{
   
		a[i].push_back(0);
		for(int j=1;j<=n;j++)
		{
   
			int temp;
			scanf("%d",&temp);
			a[i].push_back(temp);
		}
	}
	memset(f,0x3f3f3f3f,sizeof(f));
	for(int j=1;j<=n;j++)
	{
   
	for(int i=1;i<=m;i++)
	{
   
		if(i==1)
		p[i]=1;
		else
		{
   
			if(a[i][j]>=a[i-1][j])
			{
   p[i]=p[i-1];
			}
			else
			p[i]=i;
		}
		
	}
	for(int i=1;i<=m;i++)
	f[i]=min(p[i],f[i]);
	}
	int t;
	scanf("%d",&t);
	while(t--)
	{
   
		int l,r;
		scanf("%d%d",&l,&r);
		if(f[r]<=l)	printf("Yes\n");
		else	printf("No\n");
	}
}