#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int, int>;
template <typename T>
bool chmin(T& a, const T& b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T& val) {
  fill((T*)array, (T*)(array + N), val);
}
vector<int> dx = {-1, 1, 0, 0, -1, -1, 1, 1};
vector<int> dy = {0, 0, -1, 1, -1, 1, -1, 1};
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;
  vector<int> a;
  for (int i = 0; i < (k); i++) {
    int l, r;
    cin >> l >> r;
    for (int j = l; j <= r; j++) a.emplace_back(j);
  }
  sort(a.begin(), a.end());
  int m = a.size();
  vector<mint> dp(n + 1, 0);
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < (m); j++) {
      if (i - a[j] >= 0)
        dp[i] += dp[i - a[j]];
      else
        break;
    }
  }
  cout << dp[n - 1] << endl;
  return 0;
}
