1007: Pass!

const int mod = 998244353;
inline int ksm(int a, int b) {
    int res = 1;
    while(b) {
        if(b & 1)   res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}
inline int inv(int x) {
    return ksm(x, mod - 2);
}
/*
1       0
0       1
10      2
90      3
910     4
9090    5
90910   6
909090  7
9090910
90909090
909090910
fi = fi-1 * 9 + fi-2 * 10
fi + fi-1 = 10^x
if(i & 1)   fi = (fi-1 - 1) * 10
else        fi = (fi-1 + 1) * 10
1、假设x是偶数项 -> x + (x - 1) * 10 = 10^m(m & 1 == 0)
2、假设x是奇数项 -> x + (x + 1) * 10 = 10^m(m & 1 == 1)
通用即将10改为(n - 1)

TLE：
优化
1、偶数项优化：即x + (x - 1) * 10 = 100^m -> ans = 2 * m
2、奇数项优化：变为 x * inv(n - 1) + 1 + x = 100^m -> ans = 2 * m + 1

还是不行
优化：将两项放在一起 ->   显然两项都是100^m，一起求

还是不行
优化：将map -> unordered_map

ok!  评测姬真---

BGSG:*/
int log(int a, int b, int ans1, int ans2) {
    int m = sqrt(mod + 0.5), v = inv(ksm(a, m));
    unordered_map<int, int> mp;   mp[1] = 0;
    for(int i = 0, e = 1; i < m; ++ i, e = e * a % mod)  {
        if(e == ans1)   return i * 2;
        if(e == ans2)   return i * 2 + 1;
        if(!mp.count(e))    mp[e] = i;
    }
    for(int i = 0, mul = 1; i < m; ++ i, mul = mul * v % mod) {
        if(mp.count(ans1 * mul % mod))  return (i * m + mp[ans1 * mul % mod]) * 2;
        if(mp.count(ans2 * mul % mod))  return (i * m + mp[ans2 * mul % mod]) * 2 + 1;
    }
    return -1;
}
void run() {
    //printf("%lld\n", ksm(10, 196003445));
    int n = rd(), x = rd(), pre = 1;
    /*for(int i = 1; i <= 10; ++ i) {
        if(i & 1)   printf("%lld\n", pre = (pre - 1) * (n - 1));
        else        printf("%lld\n", pre = (pre + 1) * (n - 1));
    }*/
    if(x == 1)  {
        puts("0");
        return ;
    }
    if(x == 0) {
        puts("1");
        return ;
    }
    int y = (x - 1) * (n - 1) % mod;
    int ans1 = (x + y) % mod;
    //x = (x + 1) * (n - 1) % mod;
    //(prex - 1) * (n - 1) = x
    //prex = x / (n - 1) + 1
    y = (x * ksm(n - 1, mod - 2) + 1) % mod;
    int ans2 = (x + y) % mod;
    int res = log((n - 1) * (n - 1) % mod, (x + y) % mod, ans1, ans2);//偶数项
    printf("%lld\n", res % mod);
    /*x = (x + 1) * (n - 1) % mod;
    y = (x - 1) * (n - 1) % mod;
    res = log((n - 1) * (n - 1) % mod, (x + y) % mod);//奇数项
    if(res != -1) printf("%lld\n", (res * 2 % mod - 1 + mod) % mod);
    else puts("-1");*/
    return ;
}
signed main() {
    int t = rd();
    while(t--) run();
    return 0;
}