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>
#define all(a) (a).begin(),(a).end()
#define rep(i,n) for(ll i=0,i##_len=(n);i<i##_len;i++)
#define rep2(i,a,b) for(ll i=(a),i##_len=(b);i<=i##_len;i++)
#define rrep(i,n) ll i=(n);ll now=-1;while(i--&&(now++||1))
#define Yes(n) cout<<((n)?Yes:No)<<"\n"
#define vll vector<ll>
#define pb push_back

using namespace std;
using ll = long long;

static inline void ignore_ret(int x) {x++;}
template<class T> bool chmax(T&a,const T&b){if(a<b){a=b;return true;}return false;}
template<class T> bool chmin(T&a,const T&b){if(a>b){a=b;return true;}return false;}
template<class T> T gcd_vec(vector<T>v){T r=v.at(0);for(T i=1;i<v.size();i++)r=gcd(r,v.at(i));return r;}
template<class T> T lcm_vec(vector<T>v){T r=v.at(0);for(T i=1;i<v.size();i++)r=lcm(r,v.at(i));return r;}
template<class T> T factorial(T n){if(n==0)return 1;T r=1;for(T i=2;i<=n;i++)r*=i;return r;}
template<class T> T nCr(T n,T k){return factorial<T>(n)/(factorial<T>(k)*factorial<T>(n-k));}
template<class T> T dot_product(vector<T>u,vector<T>v){T r=0;for(T i=0;i<u.size();i++)r+=u.at(i)*v.at(i);return r;}
template<class T> T sum(vector<T>v){return reduce(all(v),0LL);}

// Template ends here

void solve(ll N, vector<ll> P){
    ll Min = N + 1, cnt = 0;
    rep(i, N) if (chmin(Min, P.at(i))) cnt++;
    cout << cnt << "\n";
}

int main() {
    ll N;
    ignore_ret(scanf("%lld",&N));
    vector<ll> P(N);
    for(int i = 0 ; i < N ; i++){
        ignore_ret(scanf("%lld",&P[i]));
    }
    solve(N, move(P));
    return 0;
}