CodeForces 777C Alyona and Spreadsheet【一维数组模拟二维】

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 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.

题意：n行m列k个询问，每个询问[L, R]行区间是否存在非递减的列；

分析：

思维性模拟，由于nm范围未定，我用的一维数组存储，模拟的时候(i - 1)/ m + 1来表示二维；

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
typedef long long LL;

const int MAXN = 200000;
int a[MAXN], b[MAXN], c[MAXN];

int main() {
int n, m;
scanf("%d %d", &n, &m);
for(int i = 1; i <= n * m; ++i) {
scanf("%d", &a[i]);
}
for(int j = 0; j < m; ++j) {
for(int i = 1 + j; i <= n * m; i += m) {
int L = i, R = i + m;
while(a[R] >= a[R - m]) {
R += m;
}
i = R - m;
for(int k = L; k < R; k += m) {
b[k] = i;
}
}
}
for(int i = 1; i <= n * m; i += m) {
int ans = 0;
for(int j = 0; j < m; ++j) {
ans = max((b[i + j] - 1)/ m + 1, ans);
}
c[(i - 1)/ m + 1] = ans;
}
int L, R, k;
scanf("%d", &k);
while(k--) {
scanf("%d %d", &L, &R);
if(c[L] >= R) puts("Yes");
else puts("No");
}
return 0;
}