Problem: After his wife's tragic death, Eurydice, Orpheus descended to the realm of death to see her. Reaching its gates was uneasy, but passing through them proved to be even more challenging. Mostly because of Cerberus, the three-headed hound of Hades. 

Orpheus, a famous poet, and musician plans to calm Cerberus with his poetry and safely walk past him. He created a very peculiar poem for Cerberus. It consists only of lowercase English letters. 

We call a poem's substring a palindrome if and only if it reads the same backwards and forwards. A string a is a substring of a string b if a can be obtained from b by deleting several (possibly zero or all) characters from the beginning and several (possibly zero or all) characters from the end.

Unfortunately, Cerberus dislikes palindromes of length greater than 1. For example in the poem abaa the hound of Hades wouldn't like substrings aba and aa.

Orpheus can only calm Cerberus if the hound likes his poetry. That's why he wants to change his poem so that it does not contain any palindrome substrings of length greater than 1.

Orpheus can modify the poem by replacing a letter at any position with any lowercase English letter. He can use this operation arbitrarily many times (possibly zero). Since there can be many palindromes in his poem, he may have to make some corrections. But how many, exactly? Given the poem, determine the minimal number of letters that have to be changed so that the poem does not contain any palindromes of length greater than 1.

Input

The first line of the input contains a single integer t (1 ≤ t ≤ 10^5) denoting the number of test cases, then t test cases follow.

The first and only line of each test case contains a non-empty string of lowercase English letters, Orpheus' poem.

The sum of the length of Orpheus' poems in all test cases will not exceed 10^5.

Output

You should output t lines, i-th line should contain a single integer, answer to the i-th test case.

Example

Input


7
babba
abaac
codeforces
zeroorez
abcdcba
bbbbbbb
a


Output


1
1
0
1
1
4
0

Note

In the first test case, we can replace the third character with c and obtain a palindrome-less poem bacba.

In the second test case, we can replace the third character with d and obtain a palindrome-less poem abdac.

In the third test case, the initial poem already doesn't contain any palindromes, so Orpheus doesn't need to change anything there.

Solution (in CPP):

Solution:
/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author Kein Yukiyoshi
 */

// clang-format off
//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define stoi stoll
#define Endl endl
#define itn int
#define fi first
#define se second
#define rep(i, n) for(int i=0, i##_len=(n); i<i##_len; i++)
#define rep2(i, a, b) for (int i = (int)(a), i##_len=(b); i < i##_len; i++)
#define rep3(i, a, b) for (int i = (int)(a), i##_len=(b); i >= i##_len; i--)
#define FOR(i, a) for (auto &i: a)
#define ALL(obj) begin(obj), end(obj)
#define _max(x) *max_element(ALL(x))
#define _min(x) *min_element(ALL(x))
#define _sum(x) accumulate(ALL(x), 0LL)

const int MOD   = 1000000007;
// const int MOD    = 998244353;
const int INF    = (int)(1e13 + 7);
const double EPS = 1e-8;
const double PI  = 3.14159265358979;
template <class T> using V = vector<T>;
template <class T> using VV = vector<vector<T>>;
template <class T> using VVV = vector<vector<vector<T>>>;
template <class T, class S>  using P = pair<T, S>;
template<class T> bool chmax(T &a, const T &b) {if (a < b) {a = b;return true;}return false;}
template<class T> bool chmin(T &a, const T &b) {if (b < a) {a = b;return true;}return false;}
int _ceil(int a, int b) { return (a >= 0 ? (a + (b - 1)) / b : (a - (b - 1)) / b); }
int _mod(int a) {return a >= 0 ? a % MOD : a - (MOD * _ceil(a, MOD));}
int _pow(int a, int b) {int res = 1;for (a %= MOD; b; a = a * a % MOD, b >>= 1)if (b & 1) res = res * a % MOD;return res;}
struct mint {long long x;mint(long long x = 0) : x((x % MOD + MOD) % MOD) {}mint operator-() const { return mint(-x); }mint &operator+=(const mint a) {if ((x += a.x) >= MOD) x -= MOD;return *this;}mint &operator-=(const mint a) {if ((x += MOD - a.x) >= MOD) x -= MOD;return *this;}mint &operator*=(const mint a) { (x *= a.x) %= MOD;return *this; }mint operator+(const mint a) const { return mint(*this) += a; }mint operator-(const mint a) const { return mint(*this) -= a; } mint operator*(const mint a) const { return mint(*this) *= a; }mint pow(long long t) const {if (!t) return 1;mint a = pow(t >> 1);a *= a;if (t & 1) a *= *this;return a;}mint inv() const { return pow(MOD - 2); }mint &operator/=(const mint a) { return *this *= a.inv(); }mint operator/(const mint a) const { return mint(*this) /= a; }};istream &operator>>(istream &is, mint &a) { return is >> a.x; }ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }
// clang-format on

class CCaninePoetry {
public:
  static void solve(istream &cin, ostream &cout) {
    cin.tie(nullptr);
    cout.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    int t;
    cin >> t;
    rep(_, t) {
      string S;
      cin >> S;
      V<int> s;
      int ans = 0;
      rep(i, S.size()) s.emplace_back(S[i]);
      int now = INF;
      if (s[0] == s[1]) {
        s[1] = now;
        now++;
        ans++;
      }
      rep2(i, 2, S.size()) {
        if (s[i] == s[i - 1] or s[i] == s[i - 2]) {
          s[i] = now;
          now++;
          ans++;
        }
      }
      cout << ans << endl;
    }
  }
};


signed main() {
  CCaninePoetry solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}