#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using P = pair<int, int>;
using pll = pair<ll, ll>;
using T = tuple<int, int, int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vs = vector<string>;
using vp = vector<P>;
using vt = vector<T>;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
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, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
int main() {
  int n, k;
  cin >> n >> k;
  vp lr(k);
  for (int((i)) = (int)(0); ((i)) < (int)((k)); ++((i)))
    cin >> lr[i].first >> lr[i].second;
  vector<mint> dp(n + 1);
  vl imo(n);
  dp[0] = 1;
  mint sum = 0;
  for (int((i)) = (int)(0); ((i)) < (int)((n)); ++((i))) {
    sum += dp[i];
    for (int((j)) = (int)(0); ((j)) < (int)((k)); ++((j))) {
      int l = min(lr[j].first, n);
      int r = min(lr[j].second + 1, n);
      dp[i + l] += sum;
      dp[i + r] -= sum;
    }
  }
  cout << dp[n - 1] << endl;
  return 0;
}
