D. Inversion Counting

D. Inversion Counting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A permutation of size n is an array of size n such that each integer from 1 to n occurs exactly once in this array. An inversion in a permutation p is a pair of indices (i, j) such that i > j and a_i < a_j. For example, a permutation [4, 1, 3, 2] contains 4 inversions: (2, 1), (3, 1), (4, 1), (4, 3).

You are given a permutation a of size n and m queries to it. Each query is represented by two indices l and r denoting that you have to reverse the segment [l, r] of the permutation. For example, if a = [1, 2, 3, 4] and a query l = 2, r = 4 is applied, then the resulting permutation is [1, 4, 3, 2].

After each query you have to determine whether the number of inversions is odd or even.

Input

The first line contains one integer n (1 ≤ n ≤ 1500) — the size of the permutation.

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ n) — the elements of the permutation. These integers are pairwise distinct.

The third line contains one integer m (1 ≤ m ≤ 2·10⁵) — the number of queries to process.

Then m lines follow, i-th line containing two integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) denoting that i-th query is to reverse a segment [l_i, r_i] of the permutation. All queries are performed one after another.

Output

Print m lines. i-th of them must be equal to odd if the number of inversions in the permutation after i-th query is odd, and even otherwise.

Examples

Input

3
1 2 3
2
1 2
2 3

Output

odd
even

Input

4
1 2 4 3
4
1 1
1 4
1 4
2 3

Output

odd
odd
odd
even

思路： 对【l，r】转置相当于对len/2对数进行交换，（交换奇数次奇偶性改变，交换偶数次不变） len = r - l + 1 

Code：

#include 
using namespace std;
const int AX = 1500 + 666;
int a[AX];
int main(){
	int n , q ;
	cin >> n;
	for( int i = 1 ; i <= n ; i++ ){
		cin >> a[i];
	}
	int cnt = 0;
	for( int i = 1 ; i <= n ; i++ ){
		for( int j = i + 1 ; j <= n ; j ++ ){
			if( a[j] < a[i] ){
				cnt ++ ;
			}
		}
	}
	cin >> q;
	int l , r;
	while( q-- ){
		cin >> l >> r;
		if( l == r ){
			cout << ( cnt % 2 == 0 ? "even" : "odd" ) << endl;
			continue;
		}
		int len = r - l + 1;
		len /= 2 ;
		int falg ;
		if( len % 2 == 1 ){
			if( cnt % 2 == 1 ){
				falg = 0;
				cnt = 0;
			}else{
				falg = 1;
				cnt = 1;
			}
		}else{
			if( cnt % 2 == 1 ){
				falg = 1;
				cnt = 1;
			}else{
				falg = 0;
				cnt = 0;
			}
		}
		cout << ( falg == 1 ? "odd" : "even" ) << endl;
	}
	return 0;
}