/** @xigua */
#include <cstdio>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <stack>
#include <cstring>
#include <queue>
#include <set>
#include <string>
#include <map>
#include <climits>
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef double db;
const int maxn = 1e6 + 5;
const int mod = 998244353;
const int INF = 1e8 + 5;
const ll inf = 1e15 + 5;
const db eps = 1e-6;
bool is[maxn];
ll pri[maxn]; int cnt;

void init() {
    for (int i = 2; i < maxn; i++) {
        if (!is[i]) {
            pri[++cnt] = i; //素数筛
            for (int j = i + i; j < maxn; j += i)
                is[j] = 1;
        }
    }
}

ll fac[maxn], p[maxn];

void solve() {
    ll l, r, k; cin >> l >> r >> k;
    ll res = 0;
    //通过1 到 (r - l + 1)的数组来表示 l 到 r
    //fac代表当前数的因子个数，p代表当前数被分解之后的值
    for (ll i = l; i <= r; i++)
        fac[i-l+1] = 1, p[i-l+1] = i;
    for (int i = 1; i <= cnt; i++) {
        ll be;
        if (l % pri[i] == 0) be = l;
        else {
            be = l + (pri[i] - l % pri[i]); //找到筛法的起点，因为有些不是从l开始的
        }
        for (ll j = be; j <= r; j += pri[i]) { //枚举be到r
            //每次增加pri[i]就可以保证这个数肯定是pri[i]的倍数
            ll tmp = 0;
            while (p[j-l+1] % pri[i] == 0) {    //看当前质因数的个数
                tmp++;
                p[j-l+1] /= pri[i];
            }
            fac[j-l+1] = fac[j-l+1] * (k * tmp % mod + 1) % mod;
        }
    }
    for (ll i = l; i <= r; i++) {
         //素数
        if (p[i-l+1] != 1) fac[i-l+1] = fac[i-l+1] * (k  + 1) % mod;
        res = (res + fac[i-l+1]) % mod;
    }
    cout << res << endl;
}

int main() {
    int t = 1, cas = 1;
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    init();
    scanf("%d", &t);
    while(t--) {
       // printf("Case %d: ", cas++);
        solve();
    }
    return 0;
}