codeforces777c

 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 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 ≤ 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 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.

 

本题目要求的最大[5000] [5000]的数组，普通数组存不下，故开了个vector不定长数组；

这一题是用的动态规划，起初用双循环绝对是超时的；

然后用的是动态规划，开一个b[ ][ ]存这个数字所在列，到这个数字位置的连续的单调子序列的长度；

 

以下是我的代码：

#include<cstdio>#include<cstring>#include<cstdlib>#include<vector>#include<algorithm>using namespace std; const int maxn=100005;int n,m;vector <int> a[maxn],b[maxn]; int judge(int l,int r){    int flag=0;     if(b[r][m]>=(r-l+1))        flag=1;    return flag; } int main(){    while(~scanf("%d%d",&n,&m))    {        for(int i=0; i<maxn; i++)        {            a[i].clear();            b[i].clear();        }         int x;        for(int i=0; i<n; i++)     //输入元素 a[][];            for(int j=0; j<m; j++)            {                scanf("%d",&x);                a[i].push_back(x);            }         for(int j=0; j<m; j++)     //dp得到 b[][];        {            b[0].push_back(1);            for(int i=1; i<n; i++)            {                if(a[i][j]>=a[i-1][j])                    b[i].push_back(b[i-1][j]+1);                else                    b[i].push_back(1);            }        }         for(int i=0;i<n;i++)     //把这一行递增子序列的长度存到本行最后；        {            int maxx=0;            for(int j=0;j<m;j++)            {                maxx=max(maxx,b[i][j]);            }            b[i].push_back(maxx);          }         int k;        scanf("%d",&k);        while(k--)        {            int l,r;            scanf("%d%d",&l,&r);            int flag=judge(l-1,r-1);            if(flag)  printf("Yes\n");            else  printf("No\n");        }    }    return 0;}

 

 

tag： codeforces oj

分类：动态规划，stl-vector；