#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define mem(a,b) memset((a),(b),sizeof(a))
typedef long long ll;
typedef unsigned long long ull;
#define int long long

const int mod = 235813;
const int maxn = 2e5 + 7;
typedef long long ll;
int n, arr[maxn], p[maxn];

ll fac[maxn],inv[maxn],f[maxn]; //fac[]阶乘，inv[]逆元

ll qpow(ll base, ll pow)    //快速幂 base ^ pow % mod
{
    ll ans = 1;
    while(pow)
    {
        if(pow & 1) ans = ans * base % mod;          //若不取模就去掉mod
        base = base * base % mod;
        pow >>= 1;
    }
    return ans;
}

ll C(ll n,ll m) //求组合数
{
    if(m > n || m < 0)
        return 0;
    return fac[n] * ((inv[n - m] * inv[m]) % mod) % mod;
}

void init() //预处理阶乘和逆元
{
    fac[0] = 1;
    inv[0] = 1;
    for(int i = 1;i <= maxn - 2;i++)
    {
        fac[i] = (fac[i - 1] * i) % mod;
        inv[i] = qpow(fac[i],mod - 2);
    }
}

int exgcd(int a, int b, int &d, int &x, int &y) {//d是a和b的最大公约数x,y是初始解 
	/*切记a>b,这是欧几里得算法的特征*/
	if(b == 0) {
		x = 1;
		y = 0;
		d = a;
	} else {	
		exgcd(b, a % b, d, x, y);
		/* x1=y2,y1=x2-(a/b)*y1 扩展欧几里得递归+回溯求初解的公式*/
		int t = y;
		y = x - (a / b) * y;
		x = t;
	}
	return 0;
} 

void solve() {
    cin >> n;
    for(int i = 1; i <= n; i++) {
        cin >> arr[i];
    }
    int flag = 1;
    for(int i = n; i >= 1; i--) {
        p[i] = C(n - 1, n - i) * flag;
        flag *= -1;
    }
    int sum = 0;
    for(int i = 1; i <= n; i++) {
        // cout << p[i] << '\n';
        sum = (sum + (p[i] * arr[i]) % mod) % mod;
    }
    // cout << sum << '\n';
    int ans = 0;
    for(int i = 1; i <= n; i++) {
        int temp = sum - p[i] * arr[i];
        temp = (-1 * temp + mod) % mod;
        int a, b, c, d, x, y;
		a = p[i];
		b = mod;
		c = temp;
		exgcd(a, b, d, x, y);
		if(c % d != 0) continue; 
		else{
			x *= c / d;
			y *= c / d;
			int p = b / d;
			int q = a / d;
			if(x < 0) {//将x上升最小正整数
				int k = ceil((1.0 - x) / p);
				x += p * k;
				y -= q * k;
			}else{//将x下降为最小正整数
				int k = (x - 1) / p;
				x -= p * k;
				y += q * k;
			}
			if(x != arr[i] && x >= -100000 && x <= 100000) {
				ans++;
				continue;
			} 
			if((x + mod) != arr[i] && (x + mod) >= -100000 && (x + mod) <= 100000) {
				ans++;
				continue;
			} 
			if((x - mod) != arr[i] && (x - mod) >= -100000 && (x - mod) <= 100000) {
				ans++;
				continue;
			} 
		}
    }   
    cout << ans << '\n';
}

signed main(){
    io;
    int T = 1;
    init();
    //cin >> T;
    while(T--) {
        solve();
    }
    return 0;
}