Problem:
Score : 300 points

Problem Statement

Given is a permutation P_1, \ldots, P_N of 1, \ldots, N.
Find the number of integers i (1 \leq i \leq N) that satisfy the following condition:

For any integer j (1 \leq j \leq i), P_i \leq P_j.

Constraints

1 \leq N \leq 2 \times 10^5

P_1, \ldots, P_N is a permutation of 1, \ldots, N.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
P_1 ... P_N

Output

Print the number of integers i that satisfy the condition.

Sample Input 1

5
4 2 5 1 3

Sample Output 1

3

i=1, 2, and 4 satisfy the condition, but i=3 does not - for example, P_i > P_j holds for j = 1.

Similarly, i=5 does not satisfy the condition, either. Thus, there are three integers that satisfy the condition.

Sample Input 2

4
4 3 2 1

Sample Output 2

4

All integers i (1 \leq i \leq N) satisfy the condition.

Sample Input 3

6
1 2 3 4 5 6

Sample Output 3

1

Only i=1 satisfies the condition.

Sample Input 4

8
5 7 4 2 6 8 1 3

Sample Output 4

4

Sample Input 5

1
1

Sample Output 5

1

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef double D;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef long double ld;

#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d %d",&x,&y)
#define rd3(x,y,z) scanf("%d %d %d",&x,&y,&z)
#define rl(x) scanf("%lld",&x)
#define rl2(x,y) scanf("%lld %lld",&x,&y)
#define rl3(x,y,z) scanf("%lld %lld %lld",&x,&y,&z)
#define wd(x) printf("%d" ,x)
#define wd2(x,y) printf("%d %d",x,y)
#define wl(x) printf("%lld",x)
#define wl2(x,y) printf("%lld %lld",x,y)
#define PC(x) putchar(x)
#define GC() getchar()
#define NL printf("\n")
#define SP printf(" ")
 
#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define ITER(it,a) for (__typeof(a.begin()) it = a.begin(); it != a.end(); it++)
#define FILL(a,value) memset(a, value, sizeof(a))

#define F first
#define S second
#define MP make_pair
#define PB push_back

#define SPEED ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define fastio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL)
#define fr(i,s,e) for(int i=s;i<e;i++)
#define frl(i,s,e) for(ll i=s;i<e;i++)
#define frr(i,s,e) for(int i=s-1;i>=e;i--)
#define frv(i,a) for(int i = 0;i<(int)a.size();i++)
#define frvr(i,a) for(int i = a.size()-1;i>=0;i--)
#define tr(i,a) for(typeof(a.begin()) i = a.begin(); i != a.end();i++)

#define inf 10000000000000
#define mod 1000000007
#define maxn 100009
#define maxr 100009

void solve()
{
	ll n;
	rl(n);
	VL arr(n),tmp;
	fr(i,0,n)
		rl(arr[i]);
	ll ct=0,mx=inf;

	fr(i,0,n)
	{
		mx=min(mx,arr[i]);
		if(mx==arr[i])
			ct++;
	}
	wl(ct);
	NL;
	return;
}
int main(){
	//sstd::ios::sync_with_stdio(false); cin.tie(NULL);
	solve();
	return 0;
}