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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. 思路：n*m总共100000的数据量，所以我们可以提前处理出每一行最远可以到达哪一行。不过在处理的时候需要一点剪枝。然后就离线查询了。 代码如下：

#include
   
    #define ll long longusing namespace std;const int maxx=1e5+100;vector
    
      p[maxx];int c[maxx],d[maxx];int n,m;inline int fcs(int xx,int yx,int zx){
   	if(xx>n) return n;	if(p[xx][yx]
     
      =i) continue;//如果之前行的值最远到达的行大于当前行，就直接跳过就好了，因为这一行也是到达这一行。			else c[j]=fcs(i+1,j,p[i][j]);		}		for(int j=0;j
      
       =y) puts("Yes");		else puts("No");	}}
      
     
    
   

努力加油a啊，(^o)/~

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