#include <bits/stdc++.h>
using namespace std;
const int N = 300005, mod = 998244353;
int qp(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return ans;
}
int rv[N], w[N];
void ntt(vector<int>& a, int f) {
  int n = a.size();
  for (int i = 0; i < n; i++) rv[i] = (rv[i >> 1] >> 1) | ((i & 1) * (n >> 1));
  int wn = qp(f ? (mod + 1) / 3 : 3, (mod - 1) / n);
  w[0] = 1;
  for (int i = 1; i < n; i++) w[i] = 1ll * w[i - 1] * wn % mod;
  for (int i = 0; i < n; i++)
    if (i < rv[i]) swap(a[i], a[rv[i]]);
  for (int mid = 1; mid < n; mid <<= 1) {
    for (int i = 0; i < n; i += (mid << 1)) {
      for (int j = 0; j < mid; j++) {
        int x = a[i + j],
            y = 1ll * a[i + j + mid] * w[n / (mid << 1) * j] % mod;
        a[i + j] = (x + y >= mod ? x + y - mod : x + y);
        a[i + j + mid] = (x - y < 0 ? x - y + mod : x - y);
      }
    }
  }
  if (f) {
    int iv = qp(n, mod - 2);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * iv % mod;
  }
}
vector<int> mul(vector<int> a, vector<int> b) {
  int n = 1, m = a.size() + b.size() - 1;
  while (n < m) n <<= 1;
  a.resize(n), b.resize(n);
  ntt(a, 0), ntt(b, 0);
  for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * b[i] % mod;
  ntt(a, 1);
  a.resize(m);
  return a;
}
vector<int> inv(vector<int> a) {
  int n = a.size();
  if (n == 1) {
    a[0] = qp(a[0], mod - 2);
    return a;
  }
  int m = 1;
  while (m < (n << 1)) m <<= 1;
  vector<int> a0 = a;
  a0.resize((n + 1) / 2);
  vector<int> b0 = inv(a0);
  a.resize(m), b0.resize(m);
  ntt(a, 0), ntt(b0, 0);
  for (int i = 0; i < m; i++)
    a[i] = 1ll * b0[i] * (2 - 1ll * a[i] * b0[i] % mod + mod) % mod;
  ntt(a, 1);
  a.resize(n);
  return a;
}
vector<int> qd(vector<int> a) {
  int n = a.size();
  for (int i = 1; i < n; i++) a[i - 1] = 1ll * a[i] * i % mod;
  a.pop_back();
  return a;
}
vector<int> jf(vector<int> a) {
  int n = a.size();
  a.push_back(0);
  for (int i = n; i >= 1; i--) a[i] = 1ll * a[i - 1] * qp(i, mod - 2) % mod;
  a[0] = 0;
  return a;
}
vector<int> ln(vector<int> a) {
  a = mul(qd(a), inv(a));
  return jf(a);
}
vector<int> rev(vector<int> a) {
  a[0] = 1;
  for (int i = 1; i < a.size(); i++) a[i] = (mod - a[i]) % mod;
  return a;
}
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  vector<int> a(n + 1), b(n + 1);
  int h = 1;
  for (int i = 1; i <= n; i++) {
    int t = (m + i) / 2;
    if ((m + i) & 1)
      h = 1ll * h * (t - i + 1) % mod;
    else
      h = 1ll * h * t % mod;
    h = 1ll * h * qp(i, mod - 2) % mod;
    if (i & 1)
      b[i] = (((i / 2) & 1 ? -1 : 1) * h + mod) % mod;
    else
      a[i] = (((i / 2) & 1 ? 1 : -1) * h + mod) % mod;
  }
  a = rev(a);
  auto c = mul(b, inv(a));
  c = rev(c);
  auto f = ln(inv(a)), g = ln(inv(c));
  int ans = 1ll * n * (f[n] + g[n]) % mod;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

XYMXYM and CQXYM will prepare n problems for Codeforces. The difficulty of the problem i will be an integer a_i, where a_i ≥ 0. The difficulty of the problems must satisfy a_i+a_{i+1}<m (1 ≤ i < n), and a_1+a_n<m, where m is a fixed integer. XYMXYM wants to know how many plans of the difficulty of the problems there are modulo 998 244 353.

Two plans of difficulty a and b are different only if there is an integer i (1 ≤ i ≤ n) satisfying a_i ≠ b_i.

Input

A single line contains two integers n and m (2 ≤ n ≤ 50 000, 1 ≤ m ≤ 10^9).

Output

Print a single integer — the number of different plans.

Examples

Input


3 2


Output


4


Input


5 9


Output


8105


Input


21038 3942834


Output


338529212

Note

In the first test case, the valid a are: [0,0,0], [0,0,1], [0,1,0], [1,0,0].

[1,0,1] is invalid since a_1+a_n ≥ m.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 300005, mod = 998244353;
int qp(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return ans;
}
int rv[N], w[N];
void ntt(vector<int>& a, int f) {
  int n = a.size();
  for (int i = 0; i < n; i++) rv[i] = (rv[i >> 1] >> 1) | ((i & 1) * (n >> 1));
  int wn = qp(f ? (mod + 1) / 3 : 3, (mod - 1) / n);
  w[0] = 1;
  for (int i = 1; i < n; i++) w[i] = 1ll * w[i - 1] * wn % mod;
  for (int i = 0; i < n; i++)
    if (i < rv[i]) swap(a[i], a[rv[i]]);
  for (int mid = 1; mid < n; mid <<= 1) {
    for (int i = 0; i < n; i += (mid << 1)) {
      for (int j = 0; j < mid; j++) {
        int x = a[i + j],
            y = 1ll * a[i + j + mid] * w[n / (mid << 1) * j] % mod;
        a[i + j] = (x + y >= mod ? x + y - mod : x + y);
        a[i + j + mid] = (x - y < 0 ? x - y + mod : x - y);
      }
    }
  }
  if (f) {
    int iv = qp(n, mod - 2);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * iv % mod;
  }
}
vector<int> mul(vector<int> a, vector<int> b) {
  int n = 1, m = a.size() + b.size() - 1;
  while (n < m) n <<= 1;
  a.resize(n), b.resize(n);
  ntt(a, 0), ntt(b, 0);
  for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * b[i] % mod;
  ntt(a, 1);
  a.resize(m);
  return a;
}
vector<int> inv(vector<int> a) {
  int n = a.size();
  if (n == 1) {
    a[0] = qp(a[0], mod - 2);
    return a;
  }
  int m = 1;
  while (m < (n << 1)) m <<= 1;
  vector<int> a0 = a;
  a0.resize((n + 1) / 2);
  vector<int> b0 = inv(a0);
  a.resize(m), b0.resize(m);
  ntt(a, 0), ntt(b0, 0);
  for (int i = 0; i < m; i++)
    a[i] = 1ll * b0[i] * (2 - 1ll * a[i] * b0[i] % mod + mod) % mod;
  ntt(a, 1);
  a.resize(n);
  return a;
}
vector<int> qd(vector<int> a) {
  int n = a.size();
  for (int i = 1; i < n; i++) a[i - 1] = 1ll * a[i] * i % mod;
  a.pop_back();
  return a;
}
vector<int> jf(vector<int> a) {
  int n = a.size();
  a.push_back(0);
  for (int i = n; i >= 1; i--) a[i] = 1ll * a[i - 1] * qp(i, mod - 2) % mod;
  a[0] = 0;
  return a;
}
vector<int> ln(vector<int> a) {
  a = mul(qd(a), inv(a));
  return jf(a);
}
vector<int> rev(vector<int> a) {
  a[0] = 1;
  for (int i = 1; i < a.size(); i++) a[i] = (mod - a[i]) % mod;
  return a;
}
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  vector<int> a(n + 1), b(n + 1);
  int h = 1;
  for (int i = 1; i <= n; i++) {
    int t = (m + i) / 2;
    if ((m + i) & 1)
      h = 1ll * h * (t - i + 1) % mod;
    else
      h = 1ll * h * t % mod;
    h = 1ll * h * qp(i, mod - 2) % mod;
    if (i & 1)
      b[i] = (((i / 2) & 1 ? -1 : 1) * h + mod) % mod;
    else
      a[i] = (((i / 2) & 1 ? 1 : -1) * h + mod) % mod;
  }
  a = rev(a);
  auto c = mul(b, inv(a));
  c = rev(c);
  auto f = ln(inv(a)), g = ln(inv(c));
  int ans = 1ll * n * (f[n] + g[n]) % mod;
  printf("%d\n", ans);
  return 0;
}

```