Consecutive Subsequence

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

Copy

7
3 3 4 7 5 6 8

output

Copy

4
2 3 5 6

input

Copy

6
1 3 5 2 4 6

output

Copy

2
1 4

input

Copy

4
10 9 8 7

output

Copy

1
1

input

Copy

9
6 7 8 3 4 5 9 10 11

output

Copy

6
1 2 3 7 8 9 
題解：dp[num]=dp[num-1]+1。由於num是離散的，因此能夠使用map。

 1 #pragma warning(disable:4996)
 2 #include<map>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<iostream>
 6 #include<algorithm>
 7 using namespace std;  8 #define ll long long
 9 
10 const int maxn = 200005; 11 
12 int n; 13 int a[maxn]; 14 
15 int main() 16 { 17     while (scanf("%d", &n) != EOF) { 18         int ans = 0, end; 19         map<int, int> p; 20         for (int i = 1; i <= n; i++) { 21             scanf("%d", &a[i]); 22             p[a[i]] = max(p[a[i]], p[a[i] - 1] + 1); 23             if (p[a[i]] > ans) { 24                 ans = p[a[i]]; 25                 end = a[i]; 26  } 27  } 28 
29         cout << ans << endl; 30 
31         int start = end - ans + 1; 32         for (int i = 1; i <= n; i++) { 33             if (a[i] == start) { 34                 cout << i << endl; 35                 start++; 36  } 37  } 38  } 39     return 0; 40 }