#include <iostream>
#include <deque>
#include <unordered_map>
#include <string>

using namespace std;

// 逻辑修改：将 map 放在 BFS 内部或在每次调用前清空，确保每个测试用例独立
int bfs(string& from, const string& to) {
    if (from == to) return 0;
    
    unordered_map<string, int> mp; // 记录步数
    deque<string> dq;
    
    dq.push_back(from);
    mp[from] = 0; // 初始步数为 0
    
    while (!dq.empty()) {
        string top = dq.front();
        int step = mp[top];
        dq.pop_front();
        
        if (top == to) return step;

        // 1. 尝试对每一位进行 +1 或 -1
        for (int i = 0; i < 4; i++) {
            for (int d : {1, -1}) {
                string next_s = top;
                int digit = next_s[i] - '0';
                digit += d;
                
                // 循环处理：1减1变9，9加1变1
                if (digit == 0) digit = 9;
                if (digit == 10) digit = 1;
                
                next_s[i] = digit + '0';
                if (mp.find(next_s) == mp.end()) {
                    mp[next_s] = step + 1;
                    dq.push_back(next_s);
                }
            }
        }

        // 2. 尝试交换相邻数字
        for (int i = 0; i < 3; i++) { // 4位数字有3对相邻位
            string next_s = top;
            swap(next_s[i], next_s[i+1]);
            
            if (mp.find(next_s) == mp.end()) {
                mp[next_s] = step + 1;
                dq.push_back(next_s);
            }
        }
    }
    return -1; // 理论上本题必有解
}

int main() {
    int n;
    if (!(cin >> n)) return 0;
    while (n--) {
        string a, b;
        cin >> a >> b;
        cout << bfs(a, b) << endl;
    }
    return 0;
}