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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

const max=200000;
var
        a:array[1..max] of longint;
        s,n,i,j,min,d,tam:longint;
begin
        read(n);
        for i:=1 to n do read(a[i]);
        for i:=1 to n do s:=s+a[i];
        if s=a[i]*n then begin write('-1'); exit; end else
        for i:=1 to n do
        min:=a[1];
        for i:=2 to n do if a[i]<min then min:=a[i];
        tam:=min;
        for i:=1 to n do
        begin
                if (a[i]>min) and (min+1<>tam) then inc(d)
                else if a[i]<=min then begin min:=a[i]; inc(tam);  end;
        end;
        if d>=0 then write(d) else write(-1);
end.