You are given a square board, consisting of n rows and n columns. Each tile in it should be colored either white or black.

Let's call some coloring beautiful if each pair of adjacent rows are either the same or different in every position. The same condition should be held for the columns as well.

Let's call some coloring suitable if it is beautiful and there is no rectangle of the single color, consisting of at least k tiles.

Your task is to count the number of suitable colorings of the board of the given size.

Since the answer can be very large, print it modulo 998244353.

Input

A single line contains two integers n and k (1 ≤ n ≤ 500, 1 ≤ k ≤ n^2) — the number of rows and columns of the board and the maximum number of tiles inside the rectangle of the single color, respectively.

Output

Print a single integer — the number of suitable colorings of the board of the given size modulo 998244353.

Examples

Input

1 1


Output

0


Input

2 3


Output

6


Input

49 1808


Output

359087121

Note

Board of size 1 × 1 is either a single black tile or a single white tile. Both of them include a rectangle of a single color, consisting of 1 tile.

Here are the beautiful colorings of a board of size 2 × 2 that don't include rectangles of a single color, consisting of at least 3 tiles:

<image>

The rest of beautiful colorings of a board of size 2 × 2 are the following:

<image>
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using uint = unsigned;
using pii = std::pair<int, int>;
using pll = std::pair<ll, ll>;
template <typename T>
using vt = std::vector<T>;
template <typename T>
using vvt = vt<vt<T>>;
template <typename T>
using vvvt = vt<vvt<T>>;
using vi = vt<int>;
using vvi = vvt<int>;
using vvvi = vvvt<int>;
using vl = vt<ll>;
using vvl = vvt<ll>;
using vvvl = vvvt<ll>;
using vb = vt<bool>;
using vpii = vt<pii>;
static inline void Solve();
int main(int argc, char* argv[]) {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
  };
  Solve();
  return 0;
}
constexpr ll MOD = 998244353;
constexpr int MAXN = 500;
ll dp1_buff[2][MAXN + 1][MAXN + 1];
ll dp2_buff[2][MAXN + 1];
void Solve() {
  int n, k;
  cin >> n >> k;
  vl pow2(n + 1, 1);
  for (int i = 1; i <= n; i++) {
    pow2[i] = (pow2[i - 1] * 2) % MOD;
  }
  auto dp1 = dp1_buff[0], dp1_next = dp1_buff[1];
  dp1[1][1] = 2;
  for (int k = 2; k <= n; k++) {
    for (int i = 1; i <= k; i++) {
      for (int j = 1; j <= i; j++) {
        dp1_next[i][j] = dp1[i][j - 1];
      }
      for (int j = 1; j <= i; j++) {
        dp1_next[i][1] += dp1[i][j];
      }
      dp1_next[i][i] += dp1[i - 1][i - 1];
      dp1_next[i][i] %= MOD;
      dp1_next[i][1] %= MOD;
    }
    swap(dp1, dp1_next);
  }
  ll ans = 0;
  for (int i = 1; i <= min(n, k - 1); i++) {
    const int ml = min((k - 1) / i, n);
    auto dp2 = dp2_buff[0], dp2_next = dp2_buff[1];
    fill(dp2, dp2 + ml + 1, 0);
    for (int k = 1; k <= i; k++) dp2[1] += dp1[i][k];
    dp2[1] %= MOD;
    if (ml == n) {
      ans += pow2[n - 1] * dp2[1] % MOD;
      continue;
    }
    for (int k = 2; k <= n; k++) {
      for (int j = 1; j <= ml; j++) {
        dp2_next[j] = dp2[j - 1];
      }
      for (int j = 1; j <= ml; j++) {
        dp2_next[1] += dp2[j];
      }
      dp2_next[1] %= MOD;
      swap(dp2, dp2_next);
    }
    for (int j = 1; j <= ml; j++) ans += dp2[j];
  }
  cout << (ans % MOD) << endl;
}
