Codeforces Round #401 (Div. 2)C. Alyona and Spreadsheet（暴暴暴力）

C. Alyona and Spreadsheet

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

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.
题意：给定N*M的矩阵，对于第i行第j列的数字A（i,j）往下找A(i+1,j)是否大于等于A(i,j)，如果满足就一直往下一行找，然后记录A(i,j)能够到达的行数。 有K次询问，每次询问给出L和R，问第L行到第R行是否有某列满足是严格非递减序列。
题解；暴力即可。
代码：

#include<bits/stdc++.h>#define ll long longusing namespace std;const int N=100010;vector<int>mp[N];int n,m;int  ans[N];int main(){    scanf("%d%d",&n,&m);    for(int i=1; i<=n; i++)    {        mp[i].push_back(0);        for(int j=1; j<=m; j++)        {            int x;            scanf("%d",&x);            mp[i].push_back(x);        }    }    for(int j=1;j<=m;j++)    {        for(int i=1;i<n;)        {            if(mp[i+1][j]>=mp[i][j])            {                int tmp=i;                while(tmp<n&&mp[tmp+1][j]>=mp[tmp][j]) tmp++;                for(int k=i;k<tmp;k++)                    mp[k][j]=tmp;                i=tmp;            }            else            {                mp[i][j]=i;                i++;            }        }    }    memset(ans,0,sizeof(ans));    ans[n]=n;    for(int i=1; i<n; i++)    {        int mx=mp[i][1];        for(int j=2; j<=m; j++)            mx=max(mx,mp[i][j]);        ans[i]=mx;    }    int k,l,r;    scanf("%d",&k);    while(k--)    {        scanf("%d%d",&l,&r);        if(ans[l]>=r)            printf("Yes\n");        else            printf("No\n");    }    return 0;}