#include<bits/stdc++.h> /*
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including */
using namespace std;
//using namespace __gnu_pbds;

//typedefs

typedef long long LL;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> PII;
typedef pair<double, double> PDD;
typedef pair<LL, LL> PLL;
typedef vector<PII> VII;
typedef vector<PLL> VLL;

/*
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename F, typename S>
using ordered_map = tree<F, S, less<F>, rb_tree_tag, tree_order_statistics_node_update>;
*/

//#Defines
#define rep(i,a,b)	for(i=a;i<=b;i++)
#define repR(i,a,b)	for(i=a;i>=b;i--)
//#define pb push_back
#define pb emplace_back
#define F first
#define S second
#define mp make_pair
#define all(c)		c.begin(),c.end()
#define endl '\n'
#define pf printf
#define sf scanf
//#define left __left
//#define right __right
//#define tree __tree
#define MOD 1000000007
//#define harmonic(n) 0.57721566490153286l+log(n)

#define RESET(a,b)	memset(a,b,sizeof(a))
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*(b/gcd(a,b)))
#define sqr(a) ((a) * (a))

#define optimize() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define fraction() cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);

const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const LL infLL = 9000000000000000000;


//Bit Operations
inline bool checkBit(LL n, int i) { return n&(1LL<<i); }
inline LL setBit(LL n, int i) { return n|(1LL<<i);; }
inline LL resetBit(LL n, int i) { return n&(~(1LL<<i)); }

int fx[] = {0, 0, +1, -1};
int fy[] = {+1, -1, 0, 0};
//int dx[] = {+1, 0, -1, 0, +1, +1, -1, -1};
//int dy[] = {0, +1, 0, -1, +1, -1, +1, -1};


//Inline functions

inline bool EQ(double a, double b) { return fabs(a-b) < 1e-9; }
inline bool isLeapYear(LL year) { return (year%400==0) || (year%4==0 && year%100!=0); }
inline void normal(LL &a) { a %= MOD; (a < 0) && (a += MOD); }
inline LL modMul(LL a, LL b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a*b)%MOD; }
inline LL modAdd(LL a, LL b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a+b)%MOD; }
inline LL modSub(LL a, LL b) { a %= MOD, b %= MOD; normal(a), normal(b); a -= b; normal(a); return a; }
inline LL modPow(LL b, LL p) { LL r = 1; while(p) { if(p&1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline LL modInverse(LL a) { return modPow(a, MOD-2); }
inline LL modDiv(LL a, LL b) { return modMul(a, modInverse(b)); }
inline bool isInside(PII p,LL n,LL m)	{ return (p.first>=0&&p.first<n&&p.second>=0&&p.second<m); }
inline bool isInside(PII p,LL n)		{ return (p.first>=0&&p.first<n&&p.second>=0&&p.second<n); }
inline bool isSquare(LL x) 				{ LL s = sqrt(x);	return (s*s==x); }
inline bool isFib(LL x) 				{ return isSquare(5*x*x+4)|| isSquare(5*x*x-4); }
inline bool isPowerOfTwo(LL x)			{ return ((1LL<<(LL)log2(x))==x); }

//Prime Number Generator
/*
#define mx = 1000006;
bitset <mx> mark;
vector <int> primes;

void sieve() {
    mark[0] = mark[1] = 1;
    primes.push_back(2);
    int lim = sqrt(mx * 1.0) + 2;
    for (int i = 4; i < mx; i += 2) mark[i] = 1;
    for (int i = 3; i < mx; i += 2) {
        if (!mark[i]) {
            primes.push_back(i);
            if (i <= lim)
                for (int j = i * i; j < mx; j += i)
                    mark[j] = 1;
        }
    }
}
*/
int main() {
    #ifndef ONLINE_JUDGE
        freopen("in", "r", stdin);
        freopen("out", "w", stdout);
    #endif
    optimize();
    int n; cin >> n;
    LL t[n];
    for (int i = 0; i < n; i++) cin >> t[i];
    sort(t, t + n);
    LL ttl = 0, cnt = 0;
    for (int i = 0; i < n; i++) {
        if (ttl <= t[i]) {
            cnt++;
            ttl += t[i];
        }
    }
    
    cout << cnt << endl;
}

### Prompt

Please create a solution in Cpp to the following problem:

Little girl Susie went shopping with her mom and she wondered how to improve service quality. 

There are n people in the queue. For each person we know time ti needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed. 

Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue.

Input

The first line contains integer n (1 ≤ n ≤ 105).

The next line contains n integers ti (1 ≤ ti ≤ 109), separated by spaces.

Output

Print a single number — the maximum number of not disappointed people in the queue.

Examples

Input

5
15 2 1 5 3


Output

4

Note

Value 4 is achieved at such an arrangement, for example: 1, 2, 3, 5, 15. Thus, you can make everything feel not disappointed except for the person with time 5.

### Response

```cpp
#include<bits/stdc++.h> /*
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including */
using namespace std;
//using namespace __gnu_pbds;

//typedefs

typedef long long LL;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> PII;
typedef pair<double, double> PDD;
typedef pair<LL, LL> PLL;
typedef vector<PII> VII;
typedef vector<PLL> VLL;

/*
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename F, typename S>
using ordered_map = tree<F, S, less<F>, rb_tree_tag, tree_order_statistics_node_update>;
*/

//#Defines
#define rep(i,a,b)	for(i=a;i<=b;i++)
#define repR(i,a,b)	for(i=a;i>=b;i--)
//#define pb push_back
#define pb emplace_back
#define F first
#define S second
#define mp make_pair
#define all(c)		c.begin(),c.end()
#define endl '\n'
#define pf printf
#define sf scanf
//#define left __left
//#define right __right
//#define tree __tree
#define MOD 1000000007
//#define harmonic(n) 0.57721566490153286l+log(n)

#define RESET(a,b)	memset(a,b,sizeof(a))
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*(b/gcd(a,b)))
#define sqr(a) ((a) * (a))

#define optimize() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define fraction() cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);

const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const LL infLL = 9000000000000000000;


//Bit Operations
inline bool checkBit(LL n, int i) { return n&(1LL<<i); }
inline LL setBit(LL n, int i) { return n|(1LL<<i);; }
inline LL resetBit(LL n, int i) { return n&(~(1LL<<i)); }

int fx[] = {0, 0, +1, -1};
int fy[] = {+1, -1, 0, 0};
//int dx[] = {+1, 0, -1, 0, +1, +1, -1, -1};
//int dy[] = {0, +1, 0, -1, +1, -1, +1, -1};


//Inline functions

inline bool EQ(double a, double b) { return fabs(a-b) < 1e-9; }
inline bool isLeapYear(LL year) { return (year%400==0) || (year%4==0 && year%100!=0); }
inline void normal(LL &a) { a %= MOD; (a < 0) && (a += MOD); }
inline LL modMul(LL a, LL b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a*b)%MOD; }
inline LL modAdd(LL a, LL b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a+b)%MOD; }
inline LL modSub(LL a, LL b) { a %= MOD, b %= MOD; normal(a), normal(b); a -= b; normal(a); return a; }
inline LL modPow(LL b, LL p) { LL r = 1; while(p) { if(p&1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline LL modInverse(LL a) { return modPow(a, MOD-2); }
inline LL modDiv(LL a, LL b) { return modMul(a, modInverse(b)); }
inline bool isInside(PII p,LL n,LL m)	{ return (p.first>=0&&p.first<n&&p.second>=0&&p.second<m); }
inline bool isInside(PII p,LL n)		{ return (p.first>=0&&p.first<n&&p.second>=0&&p.second<n); }
inline bool isSquare(LL x) 				{ LL s = sqrt(x);	return (s*s==x); }
inline bool isFib(LL x) 				{ return isSquare(5*x*x+4)|| isSquare(5*x*x-4); }
inline bool isPowerOfTwo(LL x)			{ return ((1LL<<(LL)log2(x))==x); }

//Prime Number Generator
/*
#define mx = 1000006;
bitset <mx> mark;
vector <int> primes;

void sieve() {
    mark[0] = mark[1] = 1;
    primes.push_back(2);
    int lim = sqrt(mx * 1.0) + 2;
    for (int i = 4; i < mx; i += 2) mark[i] = 1;
    for (int i = 3; i < mx; i += 2) {
        if (!mark[i]) {
            primes.push_back(i);
            if (i <= lim)
                for (int j = i * i; j < mx; j += i)
                    mark[j] = 1;
        }
    }
}
*/
int main() {
    #ifndef ONLINE_JUDGE
        freopen("in", "r", stdin);
        freopen("out", "w", stdout);
    #endif
    optimize();
    int n; cin >> n;
    LL t[n];
    for (int i = 0; i < n; i++) cin >> t[i];
    sort(t, t + n);
    LL ttl = 0, cnt = 0;
    for (int i = 0; i < n; i++) {
        if (ttl <= t[i]) {
            cnt++;
            ttl += t[i];
        }
    }
    
    cout << cnt << endl;
}

```