Codeforces Round #359 (Div. 2) B. Little Robber Girl's Zoo

B. Little Robber Girl's Zoo

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Robber Girl likes to scare animals in her zoo for fun. She decided to arrange the animals in a row in the order of non-decreasing height. However, the animals were so scared that they couldn't stay in the right places.

The robber girl was angry at first, but then she decided to arrange the animals herself. She repeatedly names numbersl and r such thatr - l + 1 is even. After that animals that occupy positions betweenl and r inclusively are rearranged as follows: the animal at positionl swaps places with the animal at positionl + 1, the animal l + 2 swaps with the animall + 3, ..., finally, the animal at positionr - 1 swaps with the animal r.

Help the robber girl to arrange the animals in the order of non-decreasing height. You should name at most20 000 segments, since otherwise the robber girl will become bored and will start scaring the animals again.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — number of animals in the robber girl's zoo.

The second line contains n space-separated integersa₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹), where a_i is the height of the animal occupying thei-th place.

Output

Print the sequence of operations that will rearrange the animals by non-decreasing height.

The output should contain several lines, i-th of the lines should contain two space-separated integersl_i andr_i (1 ≤ l_i < r_i ≤ n) — descriptions of segments the robber girl should name. The segments should be described in the order the operations are performed.

The number of operations should not exceed 20 000.

If the animals are arranged correctly from the start, you are allowed to output nothing.

Examples

Input

42 1 4 3

Output

1 4

Input

736 28 57 39 66 69 68

Output

1 46 7

Input

51 2 1 2 1

Output

2 53 41 41 4

Note

Note that you don't have to minimize the number of operations. Any solution that performs at most20 000 operations is allowed.

题意：给你一个序列，然后你可以选择一个偶数区间[l,r]，使得swap(l,l+1)...swap(r-1,r)，到最后要使这个序列成为一个非递减序列，求经过的步骤，没有最优解限制

思路：一阵乱搞，从头开始遍历，当两个数交换的时候，会跳过，否则一个一个走，直到序列完成，注意特判序列数为1的情况

ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)#define eps 1e-8using namespace std;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}//headint a[MAXN];int main(){    int l,r,n;scanf("%d",&n);    for(int i=1;i<=n;i++) scanf("%d",&a[i]);    if(n==1)        return 0;    int bz=1;    while(bz){        l=r=1;        for(int i=1;i<n;){            if(a[i]>a[i+1])                swap(a[i],a[i+1]),r=i+1,i+=2;            else{                if((r-l+1)%2==0)                printf("%d %d\n",l,r),l=r=i+1;                else{                    if(l==r) l=r=i+1;                    else r++;                }                i++;            }            int flag=0;            for(int j=2;j<=n;j++)                if(a[j]<a[j-1]){                    flag=1;break;            }            if(flag==0) bz=0;        }        if(l!=r&&(r-l+1)%2==0)        printf("%d %d\n",l,r);        if(bz==0) break;    }    return 0;}