description: 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.
solution: #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";
    }
  }
  