经典矩阵dp寻找递增最大长度

　　竖向寻找矩阵最大递增元素长度，因为要求至少一列为递增数列，那么每行求一下最大值就可以作为len[i]:到i行截止的最长的递增数列长度。

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.

#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;

int n, m;
const int Maxn = 100002;
vector<int>a[Maxn], dp[Maxn];
int len[Maxn];

int main()
{
cin>>n>>m;
for(int i = 1; i <= n; i ++){
a[i].resize(m + 1);
dp[i].resize(m + 1);
for(int j = 1; j <= m; j ++){
scanf("%d", &a[i][j]);
}
}
for(int j = 1; j <= m; j ++) dp[1][j] = 1;
len[1] = 1;
for(int i = 2; i <= n; i ++){
for(int j = 1; j <= m; j ++){
if(a[i][j] >= a[i - 1][j])
dp[i][j] = dp[i - 1][j] + 1;
else
dp[i][j] = 1;
}
for(int j = 1; j <= m; j ++)
len[i] = max(dp[i][j], len[i]);
}
int k, l, r;
cin>>k;
while(k --){
scanf("%d %d", &l, &r);
if(len[r] >= r - l + 1)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}