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: Wrong Answer)

function Main(input) {
  /** @type {number[]} */
  const aList = input[1].split(' ').map(Number);

  // 残る可能性のあるレンガインデックス一覧
  const iList = [];
  for (var i = 0; i < aList.length; i++) {
    if (i + 1 >= aList[i]) {
      iList.push(i);
    }
  }

  // break === 0
  var f, remains, maxRemains = 0, maxIndex = null, isRemain;
  for (var flags = (1 << iList.length) - 1; flags > 0; flags--) {
    if (flags.toString(2).match(/1/g).length < maxRemains) { continue; }

    remains = 0;
    f = flags;

    iList.some((i, j) => {
      isRemain = f % 2;

      if (isRemain) {
        if (aList[i] - 1 !== remains) {
          // 中止
          return true;
        }
        remains += isRemain;
        // 最後のレンガとなる場合
        if (f === 1) {
          maxIndex = i;
          maxRemains = remains;
          // 終了
          return true;
        }
      }

      f = f >> 1;
    });
  }

  const answer = maxIndex !== null
    ? ((maxIndex + 1) - aList[maxIndex] + (aList.length - maxIndex - 1))
    : -1;

  console.log(answer);
}

Main(require("fs").readFileSync("/dev/stdin", "utf8").trim().split("\n"));