Codeforces 777C：Alyona and Spreadsheet（预处理）

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).

题意

给出一个(n imes m)的矩阵，判断第(l)行～第(r)行中是否有一列是非递减的

思路

如果这题用暴力来写的话，时间复杂度为：(O(n imes sum^{k}_{i=1}(r_i-l_i)))，有题目可知，这个时间是肯定过不去的

所以我们可以预处理：预处理从每一行往上最高到哪一行，可以保持有至少一个非递增的序列

先处理每一列的每一个位置向上的非递增序列可以延伸到哪个位置，然后每一列的对应位置去一个最大值，即可得到该行可以向上延伸的最大位置。

每次输入(l,r)，只需判断(r)行向上的位置是否小于等于(l)即可

代码

注意$ (1 ≤ n·m ≤ 100 000)$，可以直接用vector进行存，也可以用一维数组

#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];
// 当前行能往上延伸的最高位置
int can[maxn];
// 当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
";
        else
            cout<<"Yes
";
    }
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
    #endif
    return 0;
}