題干：

You are given an integer array of length?nn.

You have to choose some subsequence of this array of maximum length such that this subsequence forms a increasing sequence of consecutive integers. In other words the required sequence should be equal to?[x,x+1,…,x+k?1][x,x+1,…,x+k?1]?for some value?xx?and length?kk.

Subsequence of an array can be obtained by erasing some (possibly zero) elements from the array. You can erase any elements, not necessarily going successively. The remaining elements preserve their order. For example, for the array?[5,3,1,2,4][5,3,1,2,4]?the following arrays are subsequences:?[3][3],?[5,3,1,2,4][5,3,1,2,4],?[5,1,4][5,1,4], but the array?[1,3][1,3]?is not.

Input

The first line of the input containing integer number?nn?(1≤n≤2?1051≤n≤2?105) — the length of the array. The second line of the input containing?nn?integer numbers?a1,a2,…,ana1,a2,…,an?(1≤ai≤1091≤ai≤109) — the array itself.

Output

On the first line print?kk?— the maximum length of the subsequence of the given array that forms an increasing sequence of consecutive integers.

On the second line print the sequence of the indices of the?any?maximum length subsequence of the given array that forms an increasing sequence of consecutive integers.

Examples

Input

7 3 3 4 7 5 6 8

Output

4 2 3 5 6

Input

6 1 3 5 2 4 6

Output

2 1 4

Input

4 10 9 8 7

Output

1 1

Input

9 6 7 8 3 4 5 9 10 11

Output

6 1 2 3 7 8 9

Note

All valid answers for the first example (as sequences of indices):

• [1,3,5,6][1,3,5,6]
• [2,3,5,6][2,3,5,6]

All valid answers for the second example:

• [1,4][1,4]
• [2,5][2,5]
• [3,6][3,6]

All valid answers for the third example:

• [1][1]
• [2][2]
• [3][3]
• [4][4]

All valid answers for the fourth example:

• [1,2,3,7,8,9]

題目大意：

找一個最長的子序列，使得這個序列的數字是連續的。要求你輸出最長子序列長度，和分別位于原序列中的位置下標，spj。

解題報告：

直接map暴力一下就好。假的dp。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define FF first #define SS second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; int n,ans,mx,a[MAX]; map<int,int> mp; int main() {cin>>n;for(int x,i = 1; i<=n; i++) {scanf("%d",&x);a[i]=x;mp[x] = mp[x-1]+1;if(mp[x] > ans) {ans = mp[x];mx=x;}}cout << ans << endl;int cur = mx-ans+1;for(int i = 1; i<=n; i++) {if(a[i] == cur) printf("%d ",i),cur++;}return 0 ; }

?

總結

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