Codeforces 777C：Alyona and Spreadsheet（預處理）

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 \(a_{i, j} (1 ≤ a_{i, j} ≤ 10^9)\).

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

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 \(3\).

題意

給出一個\(n\times m\)的矩陣，判斷第\(l\)行～第\(r\)行中是否有一列是非遞減的

思路

若是這題用暴力來寫的話，時間複雜度爲：\(O(n\times \sum^{k}_{i=1}(r_i-l_i))\)，有題目可知，這個時間是確定過不去的

因此咱們能夠預處理：預處理從每一行往上最高到哪一行，能夠保持有至少一個非遞增的序列

先處理每一列的每個位置向上的非遞增序列能夠延伸到哪一個位置，而後每一列的對應位置去一個最大值，便可獲得該行能夠向上延伸的最大位置。

每次輸入\(l,r\)，只需判斷\(r\)行向上的位置是否小於等於\(l\)便可

代碼

注意$ (1 ≤ n·m ≤ 100 000)$，能夠直接用vector進行存，也能夠用一維數組

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
vector<int>ve[maxn];
// 當前行能往上延伸的最高位置
int can[maxn];
// 當前列能往上的最高位置
int line[maxn];
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,m;
    cin>>n>>m;
    int x;
    for(int i=0;i<m;i++)
        ve[0].push_back(0);
    for(int i=1;i<=n;i++)
        for(int j=0;j<m;j++)
            cin>>x,ve[i].push_back(x);
    for(int i=1;i<=n;i++)
    {
        can[i]=i;
        for(int j=0;j<m;j++)
        {
            int now_num=ve[i][j];
            int up_num=ve[i-1][j];
            if(now_num<up_num)
                line[j]=i;
            can[i]=min(can[i],line[j]);
        }
    }
    int t;
    cin>>t;
    while(t--)
    {
        int l,r;
        cin>>l>>r;
        if(can[r]>l)
            cout<<"No\n";
        else
            cout<<"Yes\n";
    }
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
    #endif
    return 0;
}