#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("sse4.2")
using namespace std;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
void flush() { cout << flush; }
void flushln() { cout << endl; }
void println() { cout << '\n'; }
template <class T>
void print(const T &x) {
  cout << x;
}
template <class T>
void read(T &x) {
  cin >> x;
}
template <class T, class... U>
void read(T &x, U &...u) {
  read(x);
  read(u...);
}
template <class T, class... U>
void print(const T &x, const U &...u) {
  print(x);
  print(u...);
}
template <class T, class... U>
void println(const T &x, const U &...u) {
  print(x);
  println(u...);
}
template <class T, class U>
inline bool chmin(T &x, const U &y) {
  return y < x ? x = y, 1 : 0;
}
template <class T, class U>
inline bool chmax(T &x, const U &y) {
  return y > x ? x = y, 1 : 0;
}
template <class T>
inline void sort(T &a) {
  sort((a).begin(), (a).end());
}
template <class T>
inline void rsort(T &a) {
  sort((a).rbegin(), (a).rend());
}
template <class T>
inline void reverse(T &a) {
  reverse((a).begin(), (a).end());
}
template <class T, class U>
inline istream &operator>>(istream &str, pair<T, U> &p) {
  return str >> p.first >> p.second;
}
template <class T>
inline istream &operator>>(istream &str, vector<T> &a) {
  for (auto &i : a) str >> i;
  return str;
}
template <class T>
inline T sorted(T a) {
  sort(a);
  return a;
}
void solve() {
  int n;
  read(n);
  string second, t;
  read(second, t);
  vector<int> dp(1 << 3, -1000000007);
  dp[0] = 0;
  auto getmex = [](int x) { return __builtin_ctz(x + 1); };
  for (int i = 0; i < n; ++i) {
    vector<int> newdp(1 << 3, -1000000007);
    int vs = second[i] - '0';
    int vt = t[i] - '0';
    for (int was = 0; was < (1 << 3); ++was) {
      chmax(newdp[was | (1 << vs) | (1 << vt)], dp[was]);
      chmax(newdp[(1 << vs) | (1 << vt)], dp[was] + getmex(was));
    }
    dp = move(newdp);
  }
  int ans = -1000000007;
  for (int i = 0; i < (1 << 3); ++i) chmax(ans, dp[i] + getmex(i));
  println(ans);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  int t;
  read(t);
  while (t--) solve();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

A binary string is a string that consists of characters 0 and 1. A bi-table is a table that has exactly two rows of equal length, each being a binary string.

Let \operatorname{MEX} of a bi-table be the smallest digit among 0, 1, or 2 that does not occur in the bi-table. For example, \operatorname{MEX} for \begin{bmatrix} 0011\\\ 1010 \end{bmatrix} is 2, because 0 and 1 occur in the bi-table at least once. \operatorname{MEX} for \begin{bmatrix} 111\\\ 111 \end{bmatrix} is 0, because 0 and 2 do not occur in the bi-table, and 0 < 2.

You are given a bi-table with n columns. You should cut it into any number of bi-tables (each consisting of consecutive columns) so that each column is in exactly one bi-table. It is possible to cut the bi-table into a single bi-table — the whole bi-table.

What is the maximal sum of \operatorname{MEX} of all resulting bi-tables can be?

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.

The first line of the description of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of columns in the bi-table.

Each of the next two lines contains a binary string of length n — the rows of the bi-table.

It's guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case print a single integer — the maximal sum of \operatorname{MEX} of all bi-tables that it is possible to get by cutting the given bi-table optimally.

Example

Input


4
7
0101000
1101100
5
01100
10101
2
01
01
6
000000
111111


Output


8
8
2
12

Note

In the first test case you can cut the bi-table as follows:

  * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
  * \begin{bmatrix} 10\\\ 10 \end{bmatrix}, its \operatorname{MEX} is 2.
  * \begin{bmatrix} 1\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 0.
  * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
  * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.
  * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.



The sum of \operatorname{MEX} is 8.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("sse4.2")
using namespace std;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
void flush() { cout << flush; }
void flushln() { cout << endl; }
void println() { cout << '\n'; }
template <class T>
void print(const T &x) {
  cout << x;
}
template <class T>
void read(T &x) {
  cin >> x;
}
template <class T, class... U>
void read(T &x, U &...u) {
  read(x);
  read(u...);
}
template <class T, class... U>
void print(const T &x, const U &...u) {
  print(x);
  print(u...);
}
template <class T, class... U>
void println(const T &x, const U &...u) {
  print(x);
  println(u...);
}
template <class T, class U>
inline bool chmin(T &x, const U &y) {
  return y < x ? x = y, 1 : 0;
}
template <class T, class U>
inline bool chmax(T &x, const U &y) {
  return y > x ? x = y, 1 : 0;
}
template <class T>
inline void sort(T &a) {
  sort((a).begin(), (a).end());
}
template <class T>
inline void rsort(T &a) {
  sort((a).rbegin(), (a).rend());
}
template <class T>
inline void reverse(T &a) {
  reverse((a).begin(), (a).end());
}
template <class T, class U>
inline istream &operator>>(istream &str, pair<T, U> &p) {
  return str >> p.first >> p.second;
}
template <class T>
inline istream &operator>>(istream &str, vector<T> &a) {
  for (auto &i : a) str >> i;
  return str;
}
template <class T>
inline T sorted(T a) {
  sort(a);
  return a;
}
void solve() {
  int n;
  read(n);
  string second, t;
  read(second, t);
  vector<int> dp(1 << 3, -1000000007);
  dp[0] = 0;
  auto getmex = [](int x) { return __builtin_ctz(x + 1); };
  for (int i = 0; i < n; ++i) {
    vector<int> newdp(1 << 3, -1000000007);
    int vs = second[i] - '0';
    int vt = t[i] - '0';
    for (int was = 0; was < (1 << 3); ++was) {
      chmax(newdp[was | (1 << vs) | (1 << vt)], dp[was]);
      chmax(newdp[(1 << vs) | (1 << vt)], dp[was] + getmex(was));
    }
    dp = move(newdp);
  }
  int ans = -1000000007;
  for (int i = 0; i < (1 << 3); ++i) chmax(ans, dp[i] + getmex(i));
  println(ans);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  int t;
  read(t);
  while (t--) solve();
  return 0;
}

```