#include <bits/stdc++.h>
using namespace std;
long gcd_calculator(long a, long b) {
  if (b == 0) return a;
  return gcd_calculator(b, a % b);
}
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
void addEdge(vector<int> adj[], int u, int v) {
  adj[u].push_back(v);
  adj[v].push_back(u);
}
void DFSUtil(int u, vector<int> adj[], vector<bool>& visited) {
  visited[u] = true;
  cout << u << " ";
  for (int i = 0; i < adj[u].size(); i++)
    if (visited[adj[u][i]] == false) DFSUtil(adj[u][i], adj, visited);
}
void DFS(vector<int> adj[], int V) {
  vector<bool> visited(V, false);
  for (int u = 0; u < V; u++)
    if (visited[u] == false) DFSUtil(u, adj, visited);
}
struct Node {
  int data;
  struct Node* left;
  struct Node* right;
  Node(int val) {
    data = val;
    left = NULL;
    right = NULL;
  }
};
void printPostorder(struct Node* node) {
  if (node == NULL) return;
  printPostorder(node->left);
  printPostorder(node->right);
  cout << node->data << " ";
}
void printInorder(struct Node* node) {
  if (node == NULL) return;
  printInorder(node->left);
  cout << node->data << " ";
  printInorder(node->right);
}
void printPreorder(struct Node* node) {
  if (node == NULL) return;
  cout << node->data << " ";
  printPreorder(node->left);
  printPreorder(node->right);
}
void printLevelOrder(Node* root) {
  if (root == NULL) return;
  queue<Node*> q;
  q.push(root);
  while (q.empty() == false) {
    Node* node = q.front();
    cout << node->data << " ";
    q.pop();
    if (node->left != NULL) q.push(node->left);
    if (node->right != NULL) q.push(node->right);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  int n;
  string s1, s2;
  for (int i = 0; i < t; i++) {
    cin >> n;
    cin >> s1 >> s2;
    string x1 = "";
    long long ans = 0;
    for (int j = 0; j < n; j++) {
      if ((s1[j] == '1' && s2[j] == '0') || (s1[j] == '0' && s2[j] == '1')) {
        ans += 2;
        x1 += '3';
      } else {
        x1 += s1[j];
      }
    }
    int count_0 = 0;
    for (int j = 0; j < x1.length(); j++) {
      if (x1[j] == '0') {
        count_0++;
      }
    }
    int a[x1.length()];
    for (int j = 0; j < x1.length(); j++) {
      a[j] = 0;
    }
    int c1 = 0;
    for (int j = 0; j < x1.length(); j++) {
      if (x1[j] == '1') {
        if (j - 1 >= 0 && x1[j - 1] == '0' && a[j - 1] == 0) {
          c1++;
          a[j - 1] = 1;
        } else if (j + 1 < x1.length() && x1[j + 1] == '0' && a[j + 1] == 0) {
          c1++;
          a[j + 1] = 1;
        }
      }
    }
    count_0 = count_0 - c1;
    if (count_0 < 0) {
      count_0 = 0;
    }
    ans += (count_0 * 1) + (c1 * 2);
    cout << ans << "\n";
  }
}

### Prompt

Please create a solution in Cpp to the following problem:

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>
using namespace std;
long gcd_calculator(long a, long b) {
  if (b == 0) return a;
  return gcd_calculator(b, a % b);
}
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
void addEdge(vector<int> adj[], int u, int v) {
  adj[u].push_back(v);
  adj[v].push_back(u);
}
void DFSUtil(int u, vector<int> adj[], vector<bool>& visited) {
  visited[u] = true;
  cout << u << " ";
  for (int i = 0; i < adj[u].size(); i++)
    if (visited[adj[u][i]] == false) DFSUtil(adj[u][i], adj, visited);
}
void DFS(vector<int> adj[], int V) {
  vector<bool> visited(V, false);
  for (int u = 0; u < V; u++)
    if (visited[u] == false) DFSUtil(u, adj, visited);
}
struct Node {
  int data;
  struct Node* left;
  struct Node* right;
  Node(int val) {
    data = val;
    left = NULL;
    right = NULL;
  }
};
void printPostorder(struct Node* node) {
  if (node == NULL) return;
  printPostorder(node->left);
  printPostorder(node->right);
  cout << node->data << " ";
}
void printInorder(struct Node* node) {
  if (node == NULL) return;
  printInorder(node->left);
  cout << node->data << " ";
  printInorder(node->right);
}
void printPreorder(struct Node* node) {
  if (node == NULL) return;
  cout << node->data << " ";
  printPreorder(node->left);
  printPreorder(node->right);
}
void printLevelOrder(Node* root) {
  if (root == NULL) return;
  queue<Node*> q;
  q.push(root);
  while (q.empty() == false) {
    Node* node = q.front();
    cout << node->data << " ";
    q.pop();
    if (node->left != NULL) q.push(node->left);
    if (node->right != NULL) q.push(node->right);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  int n;
  string s1, s2;
  for (int i = 0; i < t; i++) {
    cin >> n;
    cin >> s1 >> s2;
    string x1 = "";
    long long ans = 0;
    for (int j = 0; j < n; j++) {
      if ((s1[j] == '1' && s2[j] == '0') || (s1[j] == '0' && s2[j] == '1')) {
        ans += 2;
        x1 += '3';
      } else {
        x1 += s1[j];
      }
    }
    int count_0 = 0;
    for (int j = 0; j < x1.length(); j++) {
      if (x1[j] == '0') {
        count_0++;
      }
    }
    int a[x1.length()];
    for (int j = 0; j < x1.length(); j++) {
      a[j] = 0;
    }
    int c1 = 0;
    for (int j = 0; j < x1.length(); j++) {
      if (x1[j] == '1') {
        if (j - 1 >= 0 && x1[j - 1] == '0' && a[j - 1] == 0) {
          c1++;
          a[j - 1] = 1;
        } else if (j + 1 < x1.length() && x1[j + 1] == '0' && a[j + 1] == 0) {
          c1++;
          a[j + 1] = 1;
        }
      }
    }
    count_0 = count_0 - c1;
    if (count_0 < 0) {
      count_0 = 0;
    }
    ans += (count_0 * 1) + (c1 * 2);
    cout << ans << "\n";
  }
}

```