逆序数 技巧题

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

31 2 321 22 3

Output

oddeven

Input

41 2 4 341 11 41 42 3

Output

oddoddoddeven

Note

The first example:

1. after the first query a?=?[2,?1,?3], inversion: (2,?1);
2. after the second query a?=?[2,?3,?1], inversions: (3,?1), (3,?2).

The second example:

1. a?=?[1,?2,?4,?3], inversion: (4,?3);
2. a?=?[3,?4,?2,?1], inversions: (3,?1), (4,?1), (3,?2), (4,?2), (4,?3);
3. a?=?[1,?2,?4,?3], inversion: (4,?3);
4. a?=?[1,?4,?2,?3], inversions: (3,?2), (4,?2).

题目分析 ： 让你求一下再经过 m 次操作后，序列中的逆序数是多少，会发现一个规律， n 个数的序列中最多有逆序数 (n - 1) * n / 2 , 那么当你交换此序列的顺序后逆序数会变成 sum - 原有的逆序数 , 然后呢 再看此题，它就是让你判断一个奇偶性 ，可以借着此规律搞搞 就出来了

代码示例 ：

#define ll long long

int pre[1505];

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int n, m;
    int l, r;

    cin >> n;
    for(int i = 1; i <= n; i++){
        scanf("%d", &pre[i]);
    }
    cin >> m;
    ll cnt = 0;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j < i; j++){
            if (pre[i] < pre[j]) cnt++;
        }
    }
    for(int i = 1; i <= m; i++){
        scanf("%d%d", &l, &r);
        int t = r - l + 1;
        int y = (t-1)*t/2;
        cnt += 1ll*y;
        if (cnt % 2 == 0) printf("even\n");
        else printf("odd\n");
    }

    return 0;
}

原文地址：https://www.cnblogs.com/ccut-ry/p/8143864.html