codeforce #401 div2 Alyona and Spreadsheet 思維題

題目：

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

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.

思路參考了：ACM進階之路

思路：

我們考慮從上向下更新，引入兩個數組a{MAXN]:記錄上一行各列的數。b[MAXN]:記錄當前這一行的每一列能達到的最上面的符合條件（非嚴格單調遞增）的下標。那麼我們在討論第3行時，就不需要考慮第一列的數是多少了，我們的兩個數組已經把信息記錄下來了。我們在引入c[MAXN],記錄每一行能達到的最上面的符合條件（非嚴格單調遞增）的下標。

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;//初始化假設第i行恰好是分界點。
        for(int j=1;j<=m;++j){
            int x;scanf("%d",&x);
            if(x<a[j])b[j]=i;//如果這一行的某一個數x小於該行上一個數a[j]，那麼這一行的這一列的數最大能達到的下標是i。換句話說b[j]記錄斷點。
            a[j]=x;//更新，作爲下一行的上一行。
            if(b[j]<c[i])c[i]=b[j];//c[i]爲該行能達到的最小的下標。
        }
    }
    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");
    }
}