#include<iostream>
#include<string>
#include<algorithm>    
#include<cmath>
#include<ctime>
#include<map>
#include<vector>
#include<math.h>
#include<stdio.h>
#include<stack>
#include<queue>
#include<tuple>
#include<cassert>
#include<set>
#include<bitset>
#include<functional>
#include <fstream>
//#include<bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#define rep(i, x) for(ll i = 0; i < x; i++)
#define rep2(i, x) for(ll i = 1; i <= x; i++)
#define rep3(i, x, y) for(ll i = x; i < y; i++)
#define rep4(i, x) for(ll i = x; i >= 0; i--)
#define all(a) (a).begin(),(a).end()
#define puts(x) cout << (x) << "\n"
using ll = long long;
using ld = long double;
using namespace std;
const ll INF = 1000000000000000000;
const int intINF = 1000000000;
const ll mod = 1000000007;
const ll MOD = 998244353;
const ld pi = 3.141592653589793238;
//const ld EPS = 1e-9;

bool isprime(int p) {
	if (p == 1) return false;
	for (int i = 2; i < p; i++) {
		if (p % i == 0) return false;
	}
	return true;
}
ll gcd(ll a, ll b) {
	if (b == 0) { return a; }
	return gcd(b, a % b);
}
//ax + by = cが整数解をもつための必要十分条件は c が gcd(a,b) で割り切れること。
// 返り値: a と b の最大公約数
// ax + by = gcd(a, b) を満たす (x, y) が格納される
//main関数内に extGCD(a, b, x, y); でx, yに解が格納
ll extGCD(ll a, ll b, ll& x, ll& y) {
	if (b == 0) {
		x = 1;
		y = 0;
		return a;
	}
	ll d = extGCD(b, a % b, y, x);
	y -= a / b * x;
	return d;
}
ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}
ll keta(ll n) {
	ll res = 0;
	while (n >= 1) {
		res += n % 10; n /= 10;
	}
	return res;
}
ll modpow(ll x, ll y, ll m) {
	ll res = 1;
	while (y) {
		if (y % 2) { res *= x; res %= m; }
		x = x * x % m; y /= 2;
	}
	return res;
}
ll kaijyo[600005];
void nCkinit(ll n, ll m) {
	ll cnt = 1; kaijyo[0] = 1;
	for (int h = 1; h <= n; h++) { cnt *= h; cnt %= m; kaijyo[h] = cnt; }
}
ll nCk(ll n, ll k, ll m) {
	/*ll a = 1, b = 1;
	for (int h = 1; h <= n; h++) { a *= h; a %= m; }
	for (int h = 1; h <= k; h++) { b *= h; b %= m; }
	for (int h = 1; h <= n - k; h++) { b *= h; b %= m; }*/
	ll a = kaijyo[n], b = kaijyo[k] * kaijyo[n - k] % m;
	return a * modpow(b, m - 2, m) % m;
}
//printf("%.10f\n", n);
typedef pair <ll, ll> P;
typedef pair <ll, string> pp;
ll dx[4] = { 1, 0, -1, 0 }, dy[4] = { 0, 1, 0, -1 };
struct edge { ll to, cost; };
struct Point {
	ll x, y;
};
bool operator<(const Point& a1, const Point& a2) {
	if (a1.x < a2.x) return true;
	if (a1.x > a2.x) return false;
	if (a1.y < a2.y) return true;
	return false;
}
struct status {
	ll c;
	ll x;
	ll y;
	ll d;

	bool operator<(const status& rhs) const { return c < rhs.c; };
	bool operator>(const status& rhs) const { return c > rhs.c; };
};



int main() {
	ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
	//cout << fixed << setprecision(15);

	//input
	ll n, cnt = 0; cin >> n;
	rep2(i, 1000000) {
		cnt += i;
		if (cnt >= n) {
			cout << i << endl; return 0;
		}
	}
	return 0;
}
#include <iostream>
#include <cstdio>
using namespace std;
int t[210000];
int main(){
	int a;scanf("%d",&a);
	int at=0;
	for(int i=0;i<a;i++){
		int s;scanf("%d",&s);
		t[s]++;
		while(t[at]){
			at++;
		}
		printf("%d\n",at);
	}
}