CodeForces - 777C Alyona and Spreadsheet （vector存储二维数组）

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 lito 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.

题目大意：给你一个矩阵的数，然后询问，从第l行到第r行，是否存在一列是不下降的数。

题目分析：因为题目开头给的矩阵只告诉你总数小于10万，但是如果开10万*10万的数组很显然是会爆的，所以很明显用vector动态存储，很尴尬，这个是盲点，因为不知道动态存储二维数组，所以后来还是翻的题解，解决了这个障碍之后就比较简单了，开一个数组，记录这一行能够到达的符合要求的最大行，然后当询问的时候，直接比较能否查这个数组记录的位置是否大于询问的那一行就行了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
using namespace std;
vector<int> a[100005];//动态存储二维数组
int f[100005];
int main(){
int n,m;
while((scanf("%d%d",&n,&m))!=EOF){
int x;
for(int i=0;i<100005;i++) f[i]=i;
//memset(f,0,sizeof(f));
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
scanf("%d",&x);
a[i].push_back(x);//vector存储二维数组
}
}
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
int k;
for(k=j+1;k<n;k++){
if(a[k][i] < a[k-1][i])
break;//找到向下不递减的最大数
}
while(j != k){
f[j]=max(f[j],k-1);//更新这一行的向下的最大值
j++; //沿途的行的最大值都可以顺便更新，可以节省时间
}
j--;
}
}
/*/ for(int i=0;i<m;i++)
printf("%d ",f[i]);
printf("\n");/*/
int t;
scanf("%d",&t);
while(t--){
int l,r;
cin>>l>>r;
if(f[l-1] >= r-1)//因为数组行和列数都是从0开始，所以l和r都-1
printf("Yes\n");
else printf("No\n");
}
}
return 0;
}