codeforce #401 div2 Alyona and Spreadsheet 思维题

题目：

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 thatai, 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

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.

思路参考了：ACM进阶之路

思路：

我们考虑从上向下更新，引入两个数组a{MAXN]:记录上一行各列的数。b[MAXN]:记录当前这一行的每一列能达到的最上面的符合条件（非严格单调递增）的下标。那么我们在讨论第3行时，就不需要考虑第一列的数是多少了，我们的两个数组已经把信息记录下来了。我们在引入c[MAXN],记录每一行能达到的最上面的符合条件（非严格单调递增）的下标。

code：

#include<cstdio>
const int MAXN=1e5+5;
int a[MAXN],b[MAXN],c[MAXN];
int main(){
int n,m;scanf("%d%d",&n,&m);
for(int i=1;i<=n;++i){
c[i]=i;//初始化假设第i行恰好是分界点。
for(int j=1;j<=m;++j){
int x;scanf("%d",&x);
if(x<a[j])b[j]=i;//如果这一行的某一个数x小于该行上一个数a[j]，那么这一行的这一列的数最大能达到的下标是i。换句话说b[j]记录断点。
a[j]=x;//更新，作为下一行的上一行。
if(b[j]<c[i])c[i]=b[j];//c[i]为该行能达到的最小的下标。
}
}
int k;scanf("%d",&k);
while(k--){
int l,r;scanf("%d%d",&l,&r);
if(c[r]<=l)printf("Yes\n");
else printf("No\n");
}
}

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