Codeforces Round #496 (Div. 3) A.Tanya and Stairways

A. Tanya and Stairways

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1,2,3,1,2,3,4$.

You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway.

The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.

Input

The first line contains $n$ ($1\le n\le 1000$) — the total number of numbers pronounced by Tanya.

The second line contains integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1,2,\dots ,x$ in that order.

The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.

Output

In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways.

Examples

input

Copy

7
1 2 3 1 2 3 4

output

Copy

2
3 4 

input

Copy

4
1 1 1 1

output

Copy

4
1 1 1 1 

input

Copy

5
1 2 3 4 5

output

Copy

1
5 

input

Copy

5
1 2 1 2 1

output

Copy

3
2 2 1 

题意：给你一个n代表上楼梯的序列，每上一个楼梯他都会查用了多少步（1,2,3,4...），让你求上了多少层楼梯，和每一层楼梯所用的步数

ACDAIMA:

#include <bits/stdc++.h>
using namespace std;
int n, a[1111], ans[1111];
int main() {
    scanf("%d", &n);
    int cnt = 0;
    for (int i = 0; i < n; i++) {
        scanf("%d", a + i);
        if (a[i] == 1 && i) ans[cnt++] = a[i - 1];
    }
    ans[cnt++] = a[n - 1];
    printf("%d\n", cnt);
    for (int i = 0; i < cnt; i++) printf("%d ", ans[i]);
}