Problem: Ksenia has an array a consisting of n positive integers a_1, a_2, …, a_n. 

In one operation she can do the following: 

  * choose three distinct indices i, j, k, and then 
  * change all of a_i, a_j, a_k to a_i ⊕ a_j ⊕ a_k simultaneously, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). 



She wants to make all a_i equal in at most n operations, or to determine that it is impossible to do so. She wouldn't ask for your help, but please, help her!

Input

The first line contains one integer n (3 ≤ n ≤ 10^5) — the length of a.

The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — elements of a.

Output

Print YES or NO in the first line depending on whether it is possible to make all elements equal in at most n operations.

If it is possible, print an integer m (0 ≤ m ≤ n), which denotes the number of operations you do.

In each of the next m lines, print three distinct integers i, j, k, representing one operation. 

If there are many such operation sequences possible, print any. Note that you do not have to minimize the number of operations.

Examples

Input


5
4 2 1 7 2


Output


YES
1
1 3 4

Input


4
10 4 49 22


Output


NO

Note

In the first example, the array becomes [4 ⊕ 1 ⊕ 7, 2, 4 ⊕ 1 ⊕ 7, 4 ⊕ 1 ⊕ 7, 2] = [2, 2, 2, 2, 2].

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long INF = 9223372036854775807;
const long long MOD = 1000000007;
const long double PI = acos(-1);
inline long long power(long long a, long long b, long long z = MOD) {
  long long res = 1;
  a = (a % z);
  while (b) {
    if (b & 1) res = (res * a) % z;
    a = (a * a) % z;
    b = b >> 1;
  }
  return res;
}
inline long long inverse(long long a, long long z = MOD) {
  return power(a, z - 2, z);
}
long long extend(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long g = extend(b, a % b, x1, y1);
  x = y1;
  y = x1 - (a / b) * y1;
  return g;
}
inline long long mat_power(long long a, long long b, long long c, long long d,
                           long long n, long long z = MOD) {
  long long a0 = 1, b0 = 0, c0 = 0, d0 = 1;
  while (n) {
    if (n & 1) {
      long long a1 = a0, b1 = b0, c1 = c0, d1 = d0;
      a0 = (a1 * a + b1 * c) % z;
      b0 = (a1 * b + b1 * d) % z;
      c0 = (c1 * a + d1 * c) % z;
      d0 = (c1 * b + d1 * d) % z;
    }
    long long a1 = a, b1 = b, c1 = c, d1 = d;
    a = (a1 * a1 + b1 * c1) % z;
    b = (a1 * b1 + b1 * d1) % z;
    c = (c1 * a1 + d1 * c1) % z;
    d = (c1 * b1 + d1 * d1) % z;
    n = n >> 1;
  }
  return c0;
}
void odd(int aray[], int n) {
  vector<pair<int, pair<int, int> > > ans;
  for (int i = 1; i + 2 <= n; i += 2)
    ans.push_back(make_pair(i, make_pair(i + 1, i + 2)));
  for (int i = 1; i <= n - 4; i += 2)
    ans.push_back(make_pair(i, make_pair(i + 1, n)));
  cout << "YES\n";
  cout << ans.size() << "\n";
  for (auto x : ans)
    cout << x.first << " " << x.second.first << " " << x.second.second << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int aray[n];
  for (int i = 0; i < n; i++) cin >> aray[i];
  if (n & 1)
    odd(aray, n);
  else {
    long long sum = 0;
    for (int i = 0; i < n; i++) sum = sum ^ aray[i];
    if (sum == 0)
      odd(aray, n - 1);
    else
      cout << "NO\n";
  }
  return 0;
}