#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int, int>;
const int INF = 1001001001;
const int takahashi = 998244353;
const int mod = 998244353;
struct mint {
  ll x;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint &operator+=(const mint a) {
    if ((x += a.x) >= mod) {
      x -= mod;
    }
    return *this;
  }
  mint &operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) {
      x -= mod;
    }
    return *this;
  }
  mint &operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const { return mint(*this) += a; }
  mint operator-(const mint a) const { return mint(*this) -= a; }
  mint operator*(const mint a) const { return mint(*this) *= a; }
  mint pow(ll t) const {
    if (!t) {
      return 1;
    }
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) {
      a *= *this;
    }
    return a;
  }
  mint inv() const { return pow(mod - 2); }
  mint &operator/=(const mint a) { return *this *= a.inv(); }
  mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream &operator>>(istream &is, mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }
ll ans = 0;
int n, k;
ll minus_counter(ll n, ll tmp) {
  ll counter = 0;
  while (1) {
    n -= tmp;
    if (n < 0) {
      return counter;
    } else {
      counter++;
    }
  }
}
int main() {
  cin >> n >> k;
  vector<int> l(k);
  vector<int> r(k);
  for (int i = 0; i < (int)(k); i++) {
    cin >> l[i];
    cin >> r[i];
  }
  vector<int> s;
  for (int i = 0; i < (int)(k); i++) {
    for (int j = l[i]; j < r[i] + 1; j++) {
      s.push_back(j);
    }
    if (l[i] == r[i]) {
      s.push_back(l[i]);
    }
  }
  sort(s.begin(), s.end());
  s.erase(std::unique(s.begin(), s.end()), s.end());
  vector<mint> dp(n + 1);
  dp[1] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < (int)(s.size()); j++) {
      if (i + j < n) {
        dp[i + s[j]] += dp[i];
      }
    }
  }
  cout << dp[n] << endl;
  return 0;
}
