#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define pb push_back

#define Yes cout << "Yes\n"
#define No cout << "No\n"
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define ff first
#define ss second
#define fori(x,y) for(int i=x;i<=(int)(y);++i)
#define forj(x,y) for(int j=x;j<=(int)(y);++j)
#define fork(x,y) for(int k=x;k<=(int)(y);++k)

#define debug(x) cerr << #x << " = " << x << endl

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

ll MOD = 998244353;
ll qpow(ll a,ll p) {ll res=1; while(p) {if (p&1) {res=res*a%MOD;} a=a*a%MOD; p>>=1;} return res;}

struct Block {
    int n, t;
    vector<int> st, ed;
    vector<int> pos, arr;
    vector<map<int,ll>> maps;
    // 上面一组长度为t，下面一组长度为n
    Block(int n): n(n) {
        int m = sqrt(n);  // 块大小
        t = n/m;  // 分块中块的数量
        if (n%m) t++;  // 不是整除，最后有一个小碎块
        
        st = ed = vector<int>(t+1, 0);
        maps = vector<map<int,ll>>(t+1, map<int,ll>());
        pos = arr = vector<int>(n+1, 0);
        fori(1, t) {
            st[i] = (i-1)*m+1;
            ed[i] = i*m;
        }
        ed[t] = n;  // 最后一块
        fori(1, n) pos[i] = (i-1)/m+1;
    }
    
    void init() {
        fori(1, n) maps[pos[i]][arr[i]]++;
    }
    /**
     * 查询区间和
     */
    map<int,ll> query(int L, int R) {
        if (L>R) return map<int,ll>();
        int p = pos[L], q = pos[R];
        
        map<int,ll> res;
        if (p == q) {
            fori(L, R) res[arr[i]]++;
            return res;
        }
        fori(p+1, q-1) for (auto [x, y]: maps[i]) res[x] += y;
        fori(L, ed[p]) res[arr[i]]++;
        fori(st[q], R) res[arr[i]]++;
        return res;
    }
};

void solve() {
	int n, m, q;
    cin >> n >> m >> q;
    vector<vector<int>> g(n+1, vector<int>());
    vector<vector<int>> rnk(m+1, vector<int>(n+1, 0));
    fori(1, m) {
        forj(1, n) cin >> rnk[i][j];
        forj(1, n-1) g[rnk[i][j]].push_back(rnk[i][j+1]);
    }

    vector<int> scc(n+1, 0), low(n+1, 0), num(n+1, 0);
    int dfn = 0, cnt = 0;  // dfn为访问序号，cnt记录强连通分量的个数和编号
    stack<int> st;  // 用于保存属于同一个强连通分量的点

    auto dfs = [&](auto self, int u) -> void {
        st.push(u);
        low[u] = num[u] = ++dfn;  // 初始化
        fori(0, g[u].size()-1) {
            int v = g[u][i];
            if (!num[v]) {  // 如果没有访问过
                self(self, v);
                low[u] = min(low[u], low[v]);  // 递归返回子节点的low
            }
            else if (!scc[v]) {  // 如果子节点v访问过且不属于某一个强连通分量（好像直接else不加判断也可以）
                low[u] = min(low[u], num[v]);  // 该点的序号
            }
        }
        if (low[u] == num[u]) {  // 栈底：scc祖先
            cnt++;  // 这是一个新的强连通分量
            while (1) {
                int v = st.top();
                st.pop();
                scc[v] = cnt;  // 记录编号
                if (v == u) break;  // 栈底为该强连通分量的祖先节点
            }
        }
    };
    fori(1, n) if (!num[i]) dfs(dfs, i);

    vector<Block> vec;
    fori(1, m) {
        Block b(n);
        forj(1, n) b.arr[j] = scc[rnk[i][j]];
        b.init();
        vec.push_back(b);
    }

    ll lastans = 0;
    while (q--) {
        int id, l, r;
        cin >> id >> l >> r;
        id = (id+lastans)%m+1;
        l = (l+lastans)%n+1;
        r = (r+lastans)%n+1;

        map<int,ll> maps = vec[id-1].query(l, r);
        ll ans = 0;
        for (auto [x, y]: maps) ans += y*(y-1)/2;
        cout << ans << endl;
        lastans = ans;
    }

}

signed main() {
	IOS;
	int t;
	cin >> t;
	while (t--) {
		solve();
	}
	return 0;
}