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 $a_{i, j}$ 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 $a_{i, j} ≤ a_{i + 1, j}$ 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 $a_{i, j} ≤ a_{i + 1, j}$ 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 $a_{i, j} (1 ≤ a_{i, j} ≤ 10^9)$.

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 $l_i$ and $r_i$ $(1 ≤ l_i ≤ r_i ≤ n)$.

Output

Print "Yes" to the $i$-th line of the output if the table consisting of rows from $l_i$ to $r_i$ 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 $3$.

The meaning of problems

Gives a \ (n \ times m \) matrix, determines whether the first \ (L \) th to \ (R & lt \) line if there is a non-decreasing of

Thinking

If this problem with violence to write, time complexity is: \ (O (n-\ Times \ SUM ^ {K} _ {I =. 1} (r_i-L_i)) \) , has a problem seen, this time is in the affirmative make life difficult

So we can Pretreatment: The pretreatment each line up from the highest to which his party, can maintain at least a non-increasing sequence

To process each column of each of the non-increasing sequence may extend upwardly to a position which position corresponds to the position of each column and then to a maximum value, to obtain the maximum position of the line may extend upwardly.

Each entry \ (L, R & lt \) , is determined simply \ (R & lt \) line up position is less than equal to \ (L \) to

Code

Note $ (1 ≤ n · m ≤ 100 000) $, can directly vectorbe stored, one-dimensional arrays may also be used

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
vector<int>ve[maxn];
// 当前行能往上延伸的最高位置
int can[maxn];
// 当前列能往上的最高位置
int line[maxn];
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,m;
    cin>>n>>m;
    int x;
    for(int i=0;i<m;i++)
        ve[0].push_back(0);
    for(int i=1;i<=n;i++)
        for(int j=0;j<m;j++)
            cin>>x,ve[i].push_back(x);
    for(int i=1;i<=n;i++)
    {
        can[i]=i;
        for(int j=0;j<m;j++)
        {
            int now_num=ve[i][j];
            int up_num=ve[i-1][j];
            if(now_num<up_num)
                line[j]=i;
            can[i]=min(can[i],line[j]);
        }
    }
    int t;
    cin>>t;
    while(t--)
    {
        int l,r;
        cin>>l>>r;
        if(can[r]>l)
            cout<<"No\n";
        else
            cout<<"Yes\n";
    }
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
    #endif
    return 0;
}

Codeforces 777C: Alyona and Spreadsheet (pre-treatment)