kjhgfx

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 9;
bool prime[maxn];
int vis[maxn]; //vis[i]表示从Q到i需要的次数
void init() {  //埃氏筛素数
    prime[1] = 1;
    for (int i = 2; i *i < maxn; i++) {
        for (int j = 2 * i; j < maxn; j += i) prime[j] = 1;
    }
}
int bfs(int Q, int W) {
    memset(vis, -1, sizeof(vis)); //用-1表示从Q到不能到i，多组数据，每次需要初始化
    queue<int> q;
    q.push(Q);
    vis[Q] = 0;  //Q到Q的次数是0
    while (q.size()) {
        int r = q.front();
        q.pop();
        if (r == W) return vis[W];
        for (int i = 0; i < 10; i++) {
            //个位
            int tm = r - r % 10 + i;
            if (prime[tm] == 0 && vis[tm] == -1 && tm >= 1000) {  //题目说改变后的数要在1000~9999范围内
                q.push(tm);
                vis[tm] = vis[r] + 1;
            }
            //十位
            tm = r - (r / 10 % 10) * 10 + i * 10;
            if (prime[tm] == 0 && vis[tm] == -1 && tm >= 1000) {
                q.push(tm);
                vis[tm] = vis[r] + 1;
            }
            //百位
            tm = r - (r / 100 % 10) * 100 + i * 100;
            if (prime[tm] == 0 && vis[tm] == -1 && tm >= 1000) {
                q.push(tm);
                vis[tm] = vis[r] + 1;
            }
            //千位
            tm = r - (r / 1000 % 10) * 1000 + i * 1000;
            if (prime[tm] == 0 && vis[tm] == -1 && tm >= 1000) {
                q.push(tm);
                vis[tm] = vis[r] + 1;
            }
        }
    }
    return 0;
}
int main() {
    init();
    int _;
    cin >> _;
    while (_--) {
        int Q, W;
        cin >> Q >> W;
        cout << bfs(Q, W) << '\n';
    }
    return 0;
}