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: Accepted)

'use strict';
function Main(inputs) {
  // 1行目がinput[0], 2行目がinput[1], …に入る
  const input = inputs.split("\n");
  const n = Number(input[0]);
  const a = input[1].split(' ').map(Number);

  let count = 0;
  let now = 1;

  for(let i = 0; i < n; i++){
    if(a[i] !== now) {
      count++;
      continue;
    }
    now++;
  }
  if(count === n){
    console.log(-1);
  }else{
    console.log(count);
  }
}
//*この行以降は編集しないでください（標準入出力から一度に読み込み、Mainを呼び出します）
Main(require("fs").readFileSync("/dev/stdin", "utf8"));