#include <bits/stdc++.h>
using namespace std;

const int md = 998244353;

inline int mul(int x, int y) {
    return 1ull * x * y % md;
}

inline int fpow(int x, int y) {
    int ans = 1;
    while (y) {
        if (y & 1) ans = mul(ans, x);
        y >>= 1; x = mul(x, x);
    }
    return ans;
}

const int N = 4005;

vector <int> vec[N];
int l[N], r[N], b[N], mxr[N], mnr[N];
int T, n, m, ans1, ans2;

int main() {
    scanf("%d", &T);
    while (T--) {
        memset(mxr, 0, sizeof(mxr));
        memset(mnr, 0x3f, sizeof(mnr));
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= m; i++) {
            scanf("%d%d", &l[i], &r[i]);
            b[i] = l[i]; b[i + m] = r[i] + 1;
        }
        for (int i = 1; i <= 2 * m; i++) vec[i].clear();
        sort(b + 1, b + 2 * m + 1);
        int len = unique(b + 1, b + 2 * m + 1) - b - 1;
        for (int i = 1; i <= m; i++) {
            l[i] = lower_bound(b + 1, b + len + 1, l[i]) - b;
            r[i] = lower_bound(b + 1, b + len + 1, r[i] + 1) - b - 1;
            vec[l[i]].push_back(r[i]);
        }
        for (int i = 1; i <= len; i++) {
            mxr[i] = mxr[i - 1];
            for (auto j : vec[i]) mxr[i] = max(mxr[i], j);
        }
        for (int i = len; i >= 1; i--) {
            mnr[i] = mnr[i + 1];
            for (auto j : vec[i]) mnr[i] = min(mnr[i], j);
        }
        ans1 = ans2 = 1;
        for (int i = 1; i <= m; i++) {
            // l[i], r[i]
            int now1 = l[i] - 1, now2 = r[i], cnt1 = 0, cnt2 = 1;
            while (1) {
                while (now1 < now2) {
                    if (mxr[now1 + 1] > now1) {
                        ++cnt1;
                        now1 = mxr[now1 + 1];
                    } else break;
                }
                if (now1 < now2) break;
                if (cnt2 * ans2 > cnt1 * ans1) {
                    ans1 = cnt2;
                    ans2 = cnt1;
                }
                if (mnr[now2 + 1] == 0x3f3f3f3f) break;
                now2 = mnr[now2 + 1]; ++cnt2;
            }
        }
        int ans = mul(ans1, fpow(ans2, md - 2));
        printf("%d\n", ans);
    }
    return 0;
}