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 C (GCC 5.4.1) (judge verdict: Accepted)


#pragma warning(disable:4996)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <float.h>

#pragma region 定義
#define i_cin(X) scanf("%d", &X)
#define i_cin2(X,Y) scanf("%d %d",&X,&Y)
#define i_cin4(X,Y,Z,W) scanf("%d %d %d %d",&X,&Y,&Z,&W)
#define i_cin3(X,Y,Z) scanf("%d %d %d",&X,&Y,&Z)
#define l_cin(X) scanf("%ld",&X)
#define f_cin(X) scanf("%f",&X)
#define l_cin2(X,Y) scanf("%ld %ld",&X,&Y)
#define s_cin(X) scanf("%s",&X)
#define ll_cin(X) scanf("%lld", &X)
#define ull_cin(X) scanf("%llu", &X)
#define lf_cin(X) scanf("%lf", &X)

#define  i_cout(X) printf("%d\n",X)
#define  f_cout(X) printf("%g\n",X)
#define  i_cout2(X,Y) printf("%d %d\n",X,Y)
#define  l_cout(X) printf("%ld\n",X)
#define  s_cout(X) printf("%s\n",X)

#define  ll_cout(X) printf("%lld\n",X)
#define  ull_cout(X) printf("%llu\n",X)



typedef long long  ll;
typedef unsigned long long  ull;

#define rept(x, s, n) for (int x = s; x < n; x++)

#define S_RTN(S)  s_cout(S);return 0


#define _itoa(A,N) sprintf(A, "%d", N);

int i_cins(int n, int* A);
int l_cins2(int n, long* A, long* B);
int s_dsort(const void* a, const void* b);
int s_asort(const void* a, const void* b);

int  _gcd(int a, int b);
int  _swp(int* a, int* b);
int _cknum(char* a, int n);
int _atoi(char* s, int len);


int s_asorts(const void* a, const void* b);

//昇順
int s_asorts(const void* a, const void* b) {
	return(strcmp((char*)a, (char*)b));
}
int s_dsort(const void* a, const void* b) {
	return(*(int*)b - *(int*)a);
}
int s_asort(const void* a, const void* b) {
	return(*(int*)a - *(int*)b);
}
int l_cins2(int n, long* a, long* b) {
	int i;
	rept(i, 0, n) {
		l_cin2(*(a + i), *(b + i));
	}
	return 0;
}
int i_cins(int n, int* a) {
	int i;
	for (i = 0; i < n; i++) {
		i_cin(*(a + i));
	}

	return 0;
}
int _gcd(int a, int b) {
	int r, tmp;
	if (a < b) { tmp = a; a = b; b = tmp; }
	r = a % b;
	while (r) {
		if (!r) break;
		a = b; b = r;
		r = a % b;
	}
	return(b);
}

long _max(long a, long b) {
	return a > b ? a : b;
}
long _min(long a, long b) {
	return a < b ? a : b;
}
int _swp(int* a, int* b)
{
	int tmp;
	tmp = *b; *b = *a; *a = tmp;
	return 0;
}

int _cknum(char* a, int n) {
	int i;
	char t = '0';

	for (i = 0; i < n; i++) {
		if (a[i] < '0' || a[i]>'9') return 1;
	}
	return 0;
}


int _atoi(char* s, int len) {
	char tmp[20];
	memcpy(tmp, s, len);
	tmp[len] = 0x00;
	return (atoi(tmp));
}

#pragma endregion
// 148-D
static int a[200001];


int main(void) {
	int n, a;
	int j, i = 1, c = 0;

	i_cin(n);
	
		for(j=0;j<n;j++) {
			i_cin(a); 
//		printf("i:%d a:%d \n", i, a);
			if (a == i) {
				i++; continue;
			}
			c++;
		}
		
	(c == n) ? i_cout(-1) : i_cout(c);
	return 0;
}
//static char s[100001];


//static char s[500000];

//int main(void) {
//	int i,n, x, y, p1 = 200000,p2;
//	char q,tmp;
//	
//	s_cin(s[p1]);
//	p2 = p1+strlen(&s[p1])-1;
//	i_cin(n);
//
//
//	for (i = 0; i < n; i++) {
//		i_cin(x);
//		if (x == 1) { tmp = s[p1]; s[p2] = s[p1]; s[p2] = tmp; }
//		i_cin(y);scanf("%c",&q); scanf("%c", &q);
//		if (y == 1) { s[p1] = q; p1--;}
//		else { s[p2] = q; p2++; }
//	}
//	s_cout(&s[p1]);
//	return 0;
//}



 //93-C
//int main(void) {
//	int a[4];
//	int i,m,cnt = 0,x[3];
//
//	i_cin3(a[0], b[1], c[2]);
//	m = _max(_max(a[0], b[1]), c[2]);
//	for (i = 0; i < 3; i++) {
//		x[i] = 0;
//		if (m == a[i]) continue;
//		x[i] = a[i] / 2;
//		cnt += a[i] % 2;
//	}
//
//	m = _max(_max(a, b), c);
//	if (m == a) { x = b; y = c;}
//	if (m == b) { x = a; y = c;}
//	if (m == c) { x = a; y = b;}
//	x = m - x; y = m - y;
//	r = x + y;
//	if (!(r%2)) {
//		i_cout(r/ 2); return 0;
//	}
//	if (y % 2) { m = y; y = x; m = x; }
//
//	cnt += x;
//	if(y > x)cnt=(y-x)/2;
//    i_cout(cnt);
//		
//	return 0;
//
//
//	return 0;
//}



//#define _N 1000000000

//int main(void)
//{
//
//	int  n,i;
//	ull   tmp=1;
//
//	i_cin(n);
//
//	for (i = 1; i <= n; i++) {
//		if (!(tmp % ((ull)_N + 7)))
//			tmp = 1;
//		tmp *= (ull)i;
//	}
//	tmp %= ((ull)_N + 7);
//		
//	ull_cout(tmp);
//	return 0;
//}