Problem: There are N cells arranged in a row, numbered 1, 2, \ldots, N from left to right.

Tak lives in these cells and is currently on Cell 1. He is trying to reach Cell N by using the procedure described below.

You are given an integer K that is less than or equal to 10, and K non-intersecting segments [L_1, R_1], [L_2, R_2], \ldots, [L_K, R_K]. Let S be the union of these K segments. Here, the segment [l, r] denotes the set consisting of all integers i that satisfy l \leq i \leq r.

* When you are on Cell i, pick an integer d from S and move to Cell i + d. You cannot move out of the cells.



To help Tak, find the number of ways to go to Cell N, modulo 998244353.

Constraints

* 2 \leq N \leq 2 \times 10^5
* 1 \leq K \leq \min(N, 10)
* 1 \leq L_i \leq R_i \leq N
* [L_i, R_i] and [L_j, R_j] do not intersect (i \neq j)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N K
L_1 R_1
L_2 R_2
:
L_K R_K


Output

Print the number of ways for Tak to go from Cell 1 to Cell N, modulo 998244353.

Examples

Input

5 2
1 1
3 4


Output

4


Input

5 2
3 3
5 5


Output

0


Input

5 1
1 2


Output

5


Input

60 3
5 8
1 3
10 15


Output

221823067

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

const int M = 998244353;

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> lr(k);
  for (int i = 0; i < k; i++) {
    cin >> lr[i].first >> lr[i].second;
  }
  vector<int> dp(n + 1);
  dp[1] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j < k; j++) {
      dp[i] = (dp[i] + dp[max(i - lr[j].first, 0)] - dp[max(i - lr[j].second - 1, 0)]) % M;
      if (dp[i] < 0) {
        dp[i] += M;
      }
    }
    if (i != n) {
      dp[i] = (dp[i] + dp[i - 1]) % M;
    }
  }
  cout << dp[n] << '\n';
  return 0;
}