Problem:
Score : 300 points

Problem Statement

Snuke has N hats. The i-th hat has an integer a_i written on it.

There are N camels standing in a circle.
Snuke will put one of his hats on each of these camels.

If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print Yes; otherwise, print No.

The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself.

What is XOR?

The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non-negative integers x_1, x_2, \ldots, x_n is defined as follows:

- When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even.

For example, 3 \oplus 5 = 6.

Constraints

All values in input are integers.

3 \leq N \leq 10^{5}

0 \leq a_i \leq 10^{9}

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 \ldots a_{N}

Output

Print the answer.

Sample Input 1

3
1 2 3

Sample Output 1

Yes

If we put the hats with 1, 2, and 3 in this order, clockwise, the condition will be satisfied for every camel, so the answer is Yes.

Sample Input 2

4
1 2 4 8

Sample Output 2

No

There is no such way to distribute the hats; the answer is No.

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

#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<string>
using namespace std;
int gcd(int a, int b) {
	int c = a % b;
	while (c != 0) {
		a = b;
		b = c;
		c = a % b;
	}
	return b;
}
struct UnionFind {
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

	UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
		for (int i = 0; i < N; i++) par[i] = i;
	}

	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}

	void unite(int x, int y) { // xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
	}

	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};
typedef long long ll;

ll M = 1000000007;

vector<ll> fac(300001); //n!(mod M)
vector<ll> ifac(300001); //k!^{M-2} (mod M)

ll mpow(ll x, ll n) {
	ll ans = 1;
	while (n != 0) {
		if (n & 1) ans = ans * x % M;
		x = x * x % M;
		n = n >> 1;
	}
	return ans;
}
void setcomb() {
	fac[0] = 1;
	ifac[0] = 1;
	for (ll i = 0; i < 300000; i++) {
		fac[i + 1] = fac[i] * (i + 1) % M; // n!(mod M)
		ifac[i + 1] = ifac[i] * mpow(i + 1, M - 2) % M; // k!^{M-2} (mod M) ←累乗にmpowを採用
	}
}
ll comb(ll a, ll b) {
	if (a == 0 && b == 0)return 1;
	if (a < b || a < 0)return 0;
	ll tmp = ifac[a - b] * ifac[b] % M;
	return tmp * fac[a] % M;
}
ll perm(ll a, ll b) {
	if (a == 0 && b == 0)return 1;
	if (a < b || a < 0)return 0;
	return fac[a] * ifac[a - b] % M;
}
// mod. m での a の逆元 a^{-1} を計算する
long long modinv(long long a) {
	long long b = M, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= M;
	if (u < 0) u += M;
	return u;
}
vector<vector<ll>> mul(vector<vector<ll>> a, vector<vector<ll>> b, int n) {
	int i, j, k, t;
	vector<vector<ll>> c(n);
	for (i = 0; i < n; i++) {
		for (j = 0; j < n; j++) {
			t = 0;
			for (k = 0; k < n; k++)
				t = (t + a[i][k] * b[k][j] % M) % M;
			c[i].push_back(t);
		}
	}
	return c;
}

template< typename Monoid >
struct SegmentTree {

	int sz;
	vector< Monoid > seg;

	const Monoid M1;

	SegmentTree(int n, const Monoid &M1) : M1(M1) {
		sz = 1;
		while (sz < n) sz <<= 1;
		seg.assign(2 * sz, M1);
	}

	void set(int k, const Monoid &x) {
		seg[k + sz] = x;
	}

	void build() {
		for (int k = sz - 1; k > 0; k--) {
			seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
		}
	}

	void update(int k, const Monoid &x) {
		k += sz;
		seg[k] = x;
		while (k >>= 1) {
			seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
		}
	}

	Monoid query(int a, int b) {
		Monoid L = M1, R = M1;
		for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
			if (a & 1) L = f(L, seg[a++]);
			if (b & 1) R = f(seg[--b], R);
		}
		return f(L, R);
	}

	Monoid value(const int &k) const {
		return seg[k + sz];
	}

	Monoid f(Monoid a, Monoid b) {
		return a + b;
	}
};

int main() {
	int n, i, a, c1 = -1, c2 = -1, c3 = -1, flag,t1,t2,t3;
	cin >> n;
	for (i = 0; i < n; i++) {
		cin >> a;
		if (c1 < 0) {
			t1 = a;
			c1 = 1;
		}
		else if (c2 < 0) {
			if (a == t1)
				c1++;
			else {
				t2 = a;
				c2 = 1;
			}
		}
		else if (c3 < 0) {
			if (a == t1)
				c1++;
			else if (a == t2)
				c2++;
			else {
				t3 = a;
				c3 = 1;
			}
		}
		else {
			if (a == t1)
				c1++;
			else if (a == t2)
				c2++;
			else if (a == t3)
				c3++;
			else {
				cout << "No" << endl;
				return 0;
			}
		}
	}
	if (t1 == 0 && c2 < 0) {
		cout << "Yes" << endl;
		return 0;
	}
	if (t1 == 0 || t2 == 0) {
		if (c3 < 0) {
			if (t1 == 0 && c1 * 2 == c2) {
				cout << "Yes" << endl;
				return 0;
			}
			if (t2 == 0 && c2 * 2 == c1) {
				cout << "Yes" << endl;
				return 0;
			}
		}
	}
	if (c1 == c2 && c2 == c3) {
		flag = t1 ^ t2;
		flag = flag = flag ^ t3;
		if (flag)
			cout << "No" << endl;
		else
			cout << "Yes" << endl;
	}
	else
		cout << "No" << endl;
}