Problem: This is the first subtask of problem F. The only differences between this and the second subtask are the constraints on the value of m and the time limit. You need to solve both subtasks in order to hack this one.

There are n+1 distinct colours in the universe, numbered 0 through n. There is a strip of paper m centimetres long initially painted with colour 0. 

Alice took a brush and painted the strip using the following process. For each i from 1 to n, in this order, she picks two integers 0 ≤ a_i < b_i ≤ m, such that the segment [a_i, b_i] is currently painted with a single colour, and repaints it with colour i. 

Alice chose the segments in such a way that each centimetre is now painted in some colour other than 0. Formally, the segment [i-1, i] is painted with colour c_i (c_i ≠ 0). Every colour other than 0 is visible on the strip.

Count the number of different pairs of sequences \\{a_i\}_{i=1}^n, \\{b_i\}_{i=1}^n that result in this configuration. 

Since this number may be large, output it modulo 998244353.

Input

The first line contains a two integers n, m (1 ≤ n ≤ 500, n = m) — the number of colours excluding the colour 0 and the length of the paper, respectively.

The second line contains m space separated integers c_1, c_2, …, c_m (1 ≤ c_i ≤ n) — the colour visible on the segment [i-1, i] after the process ends. It is guaranteed that for all j between 1 and n there is an index k such that c_k = j.

Note that since in this subtask n = m, this means that c is a permutation of integers 1 through n.

Output

Output a single integer — the number of ways Alice can perform the painting, modulo 998244353.

Examples

Input


3 3
1 2 3


Output


5


Input


7 7
4 5 1 6 2 3 7


Output


165

Note

In the first example, there are 5 ways, all depicted in the figure below. Here, 0 is white, 1 is red, 2 is green and 3 is blue.

<image>

Below is an example of a painting process that is not valid, as in the second step the segment 1 3 is not single colour, and thus may not be repainted with colour 2.

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long M7 = 1000000007LL;
inline long long getint() {
  long long _x = 0, _tmp = 1;
  char _tc = getchar();
  while ((_tc < '0' || _tc > '9') && _tc != '-') _tc = getchar();
  if (_tc == '-') _tc = getchar(), _tmp = -1;
  while (_tc >= '0' && _tc <= '9') _x *= 10, _x += (_tc - '0'), _tc = getchar();
  return _x * _tmp;
}
inline long long add(long long _x, long long _y, long long _M = M7) {
  _x += _y;
  return _x >= _M ? _x - _M : _x;
}
inline long long sub(long long _x, long long _y, long long _M = M7) {
  _x -= _y;
  return _x < 0 ? _x + _M : _x;
}
inline long long mul(long long _x, long long _y, long long _M = M7) {
  _x *= _y;
  return _x >= _M ? _x % _M : _x;
}
long long mypow(long long _a, long long _x, long long _M) {
  if (_x == 0) return 1ll;
  long long _ret = mypow(mul(_a, _a, _M), _x >> 1, _M);
  if (_x & 1) _ret = mul(_ret, _a, _M);
  return _ret;
}
long long mymul(long long _a, long long _x, long long _M) {
  if (_x == 0) return 0ll;
  long long _ret = mymul(add(_a, _a, _M), _x >> 1, _M);
  if (_x & 1) _ret = add(_ret, _a, _M);
  return _ret;
}
void sleep(double sec = 1021) {
  clock_t s = clock();
  while (clock() - s < CLOCKS_PER_SEC * sec)
    ;
}
int __ = 1, _cs;
const int mxN = 505;
const long long M = 998244353;
int n, m;
long long dp[mxN][mxN], ndp[mxN][mxN];
long long pla[mxN][mxN], prb[mxN][mxN];
int c[mxN], p[mxN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    p[c[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    dp[i][i] = dp[i][i - 1] = 1;
  }
  dp[n + 1][n] = 1;
  for (int i = n; i >= 1; i--) {
    for (int j = 1; j <= n; j++)
      for (int k = 1; k <= n; k++) pla[j][k] = prb[j][k] = ndp[j][k] = 0;
    for (int l = p[i]; l >= 1; l--) {
      for (int a = p[i]; a >= l; a--) {
        pla[l][a] = dp[l][a - 1] * dp[a][p[i] - 1] % M;
        pla[l][a] += pla[l][a + 1];
        if (pla[l][a] >= M) pla[l][a] -= M;
      }
    }
    for (int r = p[i]; r <= n; r++) {
      for (int b = p[i]; b <= r; b++) {
        prb[r][b] = dp[b + 1][r] * dp[p[i] + 1][b] % M;
        prb[r][b] += prb[r][b - 1];
        if (prb[r][b] >= M) prb[r][b] -= M;
      }
    }
    for (int l = 1; l <= p[i]; l++)
      for (int r = p[i]; r <= n; r++) dp[l][r] = pla[l][l] * prb[r][r] % M;
  }
  cout << dp[1][n] << endl;
}