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: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <utility>
#include <string>
#include <algorithm>
#include <map>
#include <vector>
#include <climits>
#include <numeric>
#include<stdio.h>
#include <queue>
#include <iomanip>
#include <float.h>
#include <set>
#include<functional>
#include <stack>
#include <time.h>
#include <climits>
#include <bitset>
#include <fstream>
#include<map>
using namespace std;
long long gcd(long long a, long long b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}
vector<long long> divisor(long long n) {
	vector< long long> ret;
	for (long long i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(begin(ret), end(ret));
	return (ret);
}
struct LazySegmentTree {
private:
	int n;
	vector<int> node, lazy;
	vector<bool> lazyFlag;

public:
	LazySegmentTree(vector<int> v) {
		int sz = (int)v.size();
		n = 1; while (n < sz) n *= 2;
		node.resize(2 * n - 1);
		lazy.resize(2 * n - 1, INT_MAX);
		lazyFlag.resize(2 * n - 1, false);

		for (int i = 0; i<sz; i++) node[i + n - 1] = v[i];
		for (int i = n - 2; i >= 0; i--) node[i] = min(node[i * 2 + 1], node[i * 2 + 2]);
	}

	void lazyEvaluate(int k, int l, int r) {
		if (lazyFlag[k]) {
			node[k] = lazy[k];
			if (r - l > 1) {
				lazy[k * 2 + 1] = lazy[k * 2 + 2] = lazy[k];
				lazyFlag[k * 2 + 1] = lazyFlag[k * 2 + 2] = true;
			}
			lazyFlag[k] = false;
		}
	}

	void update(int a, int b, int x, int k = 0, int l = 0, int r = -1) {
		if (r < 0) r = n;
		lazyEvaluate(k, l, r);
		if (b <= l || r <= a) return;
		if (a <= l && r <= b) {
			lazy[k] = x;
			lazyFlag[k] = true;
			lazyEvaluate(k, l, r);
		}
		else {
			update(a, b, x, 2 * k + 1, l, (l + r) / 2);
			update(a, b, x, 2 * k + 2, (l + r) / 2, r);
			node[k] = min(node[2 * k + 1], node[2 * k + 2]);
		}
	}

	int find(int a, int b, int k = 0, int l = 0, int r = -1) {
		if (r < 0) r = n;
		lazyEvaluate(k, l, r);
		if (b <= l || r <= a) return INT_MAX;
		if (a <= l && r <= b) return node[k];
		int vl = find(a, b, 2 * k + 1, l, (l + r) / 2);
		int vr = find(a, b, 2 * k + 2, (l + r) / 2, r);
		return min(vl, vr);
	}
};
#define ll long long
class UnionFind {
public:
	vector <ll> par; // 各元の親を表す配列
	vector <ll> siz; // 素集合のサイズを表す配列(1 で初期化)

					 // Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL) {
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.assign(sz_, 1LL);  // resize だとなぜか初期化されなかった
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}

	// Member Function
	// Find
	ll root(ll x) { // 根の検索
		while (par[x] != x) {
			x = par[x] = par[par[x]]; // x の親の親を x の親とする
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;
		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		return true;
	}

	bool issame(ll x, ll y) { // 連結判定
		return root(x) == root(y);
	}

	ll size(ll x) { // 素集合のサイズ
		return siz[root(x)];
	}
};
typedef pair<long long, long long>p;
long long mod = 1000000007;
long long calc(long long k) {
	return k*(k + 1) / 2;
}
class BIT
{
public:
	vector<int> bit;
	int M;

	BIT(int M) :
		bit(vector<int>(M + 1, 0)), M(M) {}

	int sum(int i) {
		if (!i) return 0;
		if (i == 0)return 0;
		return bit[i] + sum(i - (i&-i));
	}

	void add(int i, int x) {
		if (i > M) return;
		bit[i] += x;
		add(i + (i&-i), x);
	}
};
int n, m;
signed main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(16);
	cin >> n;
	int a[100004];
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	sort(a, a + n);
	vector<p>vec;
	vec.push_back(p(a[0], 1));
	for (int i = 1; i < n; i++) {
		if (a[i] == vec[vec.size()-1].first) {
			vec[vec.size() - 1].second += 1;
		}
		else {
			vec.push_back(p(a[i], 1));
		}
	}
	if (vec.size() == 1) {
		if (vec[0].first == 0) {
			cout << "Yes";
		}
		else {
			cout << "No";
		}
	}
	if (vec.size() == 2) {
		if (vec[0].first == 0 && (vec[0].second*2==vec[1].second)) {
			cout << "Yes";
		}
		else cout << "No";
	}
	if (vec.size() == 3) {
		if (vec[0].first!=0&&vec[0].first^vec[1].first == vec[2].first&&vec[0].second == vec[1].second&&vec[1].second == vec[2].second) {
			cout << "Yes";
		}
		else cout << "No";
	}
	if (vec.size() >= 4) {
		cout << "No";
	}
}