Problem:
Score : 400 points

Problem Statement

We have N bricks arranged in a row from left to right.

The i-th brick from the left (1 \leq i \leq N) has an integer a_i written on it.

Among them, you can break at most N-1 bricks of your choice.

Let us say there are K bricks remaining. Snuke will be satisfied if, for each integer i (1 \leq i \leq K), the i-th of those brick from the left has the integer i written on it.

Find the minimum number of bricks you need to break to satisfy Snuke's desire. If his desire is unsatisfiable, print -1 instead.

Constraints

All values in input are integers.

1 \leq N \leq 200000

1 \leq a_i \leq N

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum number of bricks that need to be broken to satisfy Snuke's desire, or print -1 if his desire is unsatisfiable.

Sample Input 1

3
2 1 2

Sample Output 1

1

If we break the leftmost brick, the remaining bricks have integers 1 and 2 written on them from left to right, in which case Snuke will be satisfied.

Sample Input 2

3
2 2 2

Sample Output 2

-1

In this case, there is no way to break some of the bricks to satisfy Snuke's desire.

Sample Input 3

10
3 1 4 1 5 9 2 6 5 3

Sample Output 3

7

Sample Input 4

1
1

Sample Output 4

0

There may be no need to break the bricks at all.

Solution in JavaScript (node.js v5.12) (judge verdict: Time Limit Exceeded)

process.stdin.resume();
process.stdin.setEncoding('utf8');

var lines = [];
var reader = require('readline').createInterface({
  input: process.stdin,
  output: process.stdout
});
reader.on('line', (line) => {
  lines.push(line);
});
reader.on('close', () => {
  var count = lines[0]
  var ary = lines[1].split(" ").map( x => Number(x) )
  var break_count = 0
  // 1から開始
  var tmp_count = 1
  var c = 0
  for(var i = 0; i < count; i++) {
    if (ary[tmp_count - 1] == tmp_count) {
      c+=tmp_count
      tmp_count++
    } else {
      break_count++
      ary.splice(tmp_count - 1, 1)
    }
  }
  // 順列になっているか
  var judge = false
  if (ary.length == 0) {
    judge = true
  }
  if (sum(ary.length) != c) {
    judge = true
  }

  // 順列になっているかつ 0 の時は0
  if (break_count == 0 && judge) {
    console.log(0)
  } else {
    if (judge) {
      console.log(-1)
    } else {
      console.log(break_count)
    }
  }
  
  // 1~nの総和を求める関数
  function sum(n) {
    return (n * (n + 1) * 0.5)
  }
});