Problem:
Score : 200 points

Problem Statement

There are N mountains ranging from east to west, and an ocean to the west.

At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.

The height of the i-th mountain from the west is H_i.

You can certainly see the ocean from the inn at the top of the westmost mountain.

For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.

From how many of these N inns can you see the ocean?

Constraints

All values in input are integers.

1 \leq N \leq 20

1 \leq H_i \leq 100

Input

Input is given from Standard Input in the following format:

N
H_1 H_2 ... H_N

Output

Print the number of inns from which you can see the ocean.

Sample Input 1

4
6 5 6 8

Sample Output 1

3

You can see the ocean from the first, third and fourth inns from the west.

Sample Input 2

5
4 5 3 5 4

Sample Output 2

3

Sample Input 3

5
9 5 6 8 4

Sample Output 3

1

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

#include <iostream>
#include<vector>
#include<map>
#include<algorithm>
#include<cstring>
#include<string>
#include <set>
#include <cassert>
#include<cmath>
#include<climits>
#include<iomanip>
using namespace std;
#define MOD 1000000007
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define ll long long
const long long INF = 1LL << 60;


template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

class UnionFind {
	vector<int> data;
public:
	UnionFind(int size) : data(size, -1) { }
	bool unionSet(int x, int y) {
		x = root(x); y = root(y);
		if (x != y) {
			if (data[y] < data[x]) swap(x, y);
			data[x] += data[y]; data[y] = x;
		}
		return x != y;
	}
	bool findSet(int x, int y) {
		return root(x) == root(y);
	}
	int root(int x) {
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	int size(int x) {
		return -data[root(x)];
	}
};

int find(vector<long long> &arr, long long obj) {
	if (arr[arr.size() - 1] < obj)return arr.size()-1;
	if (arr[0] > obj)return -1;

	int min = 0; int max = arr.size() - 1;
	for (int i = 0; i < arr.size(); i++) {
		if (arr[min+(max-min) / 2] > obj) {
			max = min + (max - min) / 2;
		}
		else {
			min = min + (max - min) / 2;
		}
		if (max - min == 1) {
			return min;
		}
	}


}

//約数求める //約数
void divisor(ll n,vector<ll> &ret) {
	for (ll i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(ret.begin(), ret.end());
	
}
//階乗
long long factorial(long long n, long long mod) {
	long long ret = 1;
	for (long long i = 1; i <= n; i++) {
		ret = (ret * i) % mod;
	}
	return ret;
}
//モジュラ逆数
long long inv_mod(long long n, long long mod) {
	long long a = n % mod, b = mod - 2, ret = 1;
	while (b > 0) {
		if (b & 1) ret = (ret * a) % mod;
		a = (a * a) % mod;
		b >>= 1;
	}
	return ret;
}

long long nPr(long long n, long long r, long long mod) {
	long long ret = 1;
	for (long long i = n; i >= n - r + 1; i--) {
		ret = ret * (i % mod) % mod;
	}
	return ret;
}

long long nCr(long long n, long long r, long long mod) {
	return nPr(n, r, mod) * inv_mod(factorial(r, mod), mod) % mod;
}

bool compare_by_b(pair<ll, ll> a, pair<ll, ll> b) {//降順second
	if (a.second != b.second) {
		return a.second > b.second;
	}
	else {
		return a.first > b.first;
	}
}

bool compare_by_a(pair<ll, ll> a, pair<ll, ll> b) {//降順first
	if (a.first != b.first) {
		return a.first > b.first;
	}
	else {
		return a.second > b.second;
	}
}

ll gcd(ll m, ll n){
	if (n == 0)return m;
	return gcd(n, m%n);
}//gcd


int main() {
	ll n;
	cin >> n;
	vector<ll> h(n);
	rep(i, n)cin >> h[i];
	ll max = 0;
	ll cnt = 0;
	rep(i, n) {
		if (max <= h[i]) {
			max = h[i];
			cnt++;
		}
	}
	cout << cnt<<endl;
	return 0;
}