Codeforces Round #401(Div. 2)C. Alyona and Spreadsheet【思维】

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本文介绍了一种针对矩阵查询的高效算法，通过预处理得到一个特殊数组，使得每次查询区间内是否存在非递减排列变得简单快捷。

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

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N*M的矩阵，一共有k次查询，每次查询区间【L,R】表示询问是否在从第L行到第R行存在某一列是非递减存在的（从上到下）。

思路：

1、看问题寻本质，观察到N*M最大不超过1e5.那么我们就从这1e5着手。

我们可以直接O（M*N）每一次枚举一列，处理出来一个数组tmp【i】=x，表示这一列中，从第x行到第i行是非递减存在的。

那么我们过程维护一个最大tmp【i】作为pre【i】=x，表示第x行到第i行是满足某一列是非递减存在的。

2、那么对应查询的时候，只要pre【R】<=L即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
vector<int >a[100500];
int pre[100500];
int ans[100500];
int tmp[100500];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)
        {
            a[i].push_back(0);
            for(int j=1;j<=m;j++)
            {
                int x;
                scanf("%d",&x);
                a[i].push_back(x);
            }
        }
        memset(pre,0x3f3f3f3f,sizeof(pre));
        for(int j=1;j<=m;j++)
        {
            for(int i=1;i<=n;i++)
            {
                if(i==1)
                {
                    tmp[i]=1;
                }
                else
                {
                    if(a[i][j]>=a[i-1][j])tmp[i]=tmp[i-1];
                    else tmp[i]=i;
                }
            }
            for(int i=1;i<=n;i++)
            {
                pre[i]=min(pre[i],tmp[i]);
            }
        }
        int q;
        scanf("%d",&q);
        while(q--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            if(pre[r]<=l)printf("Yes\n");
            else printf("No\n");
        }
    }
}