#include <bits/stdc++.h>
using namespace std;
constexpr int MAXM = 1e5 + 228;
constexpr int MOD = 998244353;
int lp[MAXM];
array<int, MAXM> dplast;
array<int, MAXM> dp;
int n;
int f(int m, vector<pair<int, int>>& lr) {
  for (int i = 0; i <= m; ++i) {
    dplast[i] = 1;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < min(lr[i].first, m + 1); ++j) dp[j] = 0;
    for (int j = lr[i].first; j <= m; ++j) {
      dp[j] = dplast[j - lr[i].first];
      if (j - lr[i].second - 1 >= 0) {
        dp[j] -= dplast[j - lr[i].second - 1];
      }
      if (dp[j] < 0) dp[j] += MOD;
    }
    dplast[0] = dp[0];
    for (int j = 1; j <= m; ++j) {
      dplast[j] = dplast[j - 1] + dp[j];
      if (dplast[j] >= MOD) dplast[j] -= MOD;
    }
  }
  return dplast[m];
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  vector<int> sim;
  for (int i = 2; i <= MAXM; ++i) {
    if (lp[i] == 0) {
      lp[i] = i;
      sim.push_back(i);
    }
    for (int j = 0; j < sim.size() && sim[j] <= lp[i] && i * sim[j] <= MAXM;
         ++j) {
      lp[sim[j] * i] = sim[j];
    }
  }
  int m;
  cin >> n >> m;
  vector<pair<int, int>> lr(n);
  for (int i = 0; i < n; ++i) {
    cin >> lr[i].first >> lr[i].second;
  }
  int ans = 0;
  vector<pair<int, int>> lr2(n);
  for (int d = 1; d <= m; ++d) {
    int last = -1;
    int cnt = 0;
    int c = d;
    while (c > 1) {
      if (lp[c] == last) {
        cnt = -1;
        break;
      }
      last = lp[c];
      c /= last;
      ++cnt;
    }
    if (cnt == -1) continue;
    for (int i = 0; i < n; ++i) {
      lr2[i] = make_pair((lr[i].first + d - 1) / d, lr[i].second / d);
    }
    int add = f(m / d, lr2);
    if (cnt & 1) {
      ans -= add;
      if (ans < 0) ans += MOD;
    } else {
      ans += add;
      if (ans >= MOD) ans -= MOD;
    }
  }
  cout << ans;
  return 0;
}

### Prompt

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

Mocha wants to be an astrologer. There are n stars which can be seen in Zhijiang, and the brightness of the i-th star is a_i. 

Mocha considers that these n stars form a constellation, and she uses (a_1,a_2,…,a_n) to show its state. A state is called mathematical if all of the following three conditions are satisfied:

  * For all i (1≤ i≤ n), a_i is an integer in the range [l_i, r_i].
  * ∑  _{i=1} ^ n a_i ≤ m.
  * \gcd(a_1,a_2,…,a_n)=1.



Here, \gcd(a_1,a_2,…,a_n) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers a_1,a_2,…,a_n.

Mocha is wondering how many different mathematical states of this constellation exist. Because the answer may be large, you must find it modulo 998 244 353.

Two states (a_1,a_2,…,a_n) and (b_1,b_2,…,b_n) are considered different if there exists i (1≤ i≤ n) such that a_i ≠ b_i.

Input

The first line contains two integers n and m (2 ≤ n ≤ 50, 1 ≤ m ≤ 10^5) — the number of stars and the upper bound of the sum of the brightness of stars.

Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ m) — the range of the brightness of the i-th star.

Output

Print a single integer — the number of different mathematical states of this constellation, modulo 998 244 353.

Examples

Input


2 4
1 3
1 2


Output


4

Input


5 10
1 10
1 10
1 10
1 10
1 10


Output


251

Input


5 100
1 94
1 96
1 91
4 96
6 97


Output


47464146

Note

In the first example, there are 4 different mathematical states of this constellation:

  * a_1=1, a_2=1.
  * a_1=1, a_2=2.
  * a_1=2, a_2=1.
  * a_1=3, a_2=1.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
constexpr int MAXM = 1e5 + 228;
constexpr int MOD = 998244353;
int lp[MAXM];
array<int, MAXM> dplast;
array<int, MAXM> dp;
int n;
int f(int m, vector<pair<int, int>>& lr) {
  for (int i = 0; i <= m; ++i) {
    dplast[i] = 1;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < min(lr[i].first, m + 1); ++j) dp[j] = 0;
    for (int j = lr[i].first; j <= m; ++j) {
      dp[j] = dplast[j - lr[i].first];
      if (j - lr[i].second - 1 >= 0) {
        dp[j] -= dplast[j - lr[i].second - 1];
      }
      if (dp[j] < 0) dp[j] += MOD;
    }
    dplast[0] = dp[0];
    for (int j = 1; j <= m; ++j) {
      dplast[j] = dplast[j - 1] + dp[j];
      if (dplast[j] >= MOD) dplast[j] -= MOD;
    }
  }
  return dplast[m];
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  vector<int> sim;
  for (int i = 2; i <= MAXM; ++i) {
    if (lp[i] == 0) {
      lp[i] = i;
      sim.push_back(i);
    }
    for (int j = 0; j < sim.size() && sim[j] <= lp[i] && i * sim[j] <= MAXM;
         ++j) {
      lp[sim[j] * i] = sim[j];
    }
  }
  int m;
  cin >> n >> m;
  vector<pair<int, int>> lr(n);
  for (int i = 0; i < n; ++i) {
    cin >> lr[i].first >> lr[i].second;
  }
  int ans = 0;
  vector<pair<int, int>> lr2(n);
  for (int d = 1; d <= m; ++d) {
    int last = -1;
    int cnt = 0;
    int c = d;
    while (c > 1) {
      if (lp[c] == last) {
        cnt = -1;
        break;
      }
      last = lp[c];
      c /= last;
      ++cnt;
    }
    if (cnt == -1) continue;
    for (int i = 0; i < n; ++i) {
      lr2[i] = make_pair((lr[i].first + d - 1) / d, lr[i].second / d);
    }
    int add = f(m / d, lr2);
    if (cnt & 1) {
      ans -= add;
      if (ans < 0) ans += MOD;
    } else {
      ans += add;
      if (ans >= MOD) ans -= MOD;
    }
  }
  cout << ans;
  return 0;
}

```