#include <iostream>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <tuple>
#include <cmath>
#include <numeric>
#include <functional>
#include <cassert>
#include <atcoder/modint>
#include <atcoder/convolution>

#define debug_value(x) cerr << "line" << __LINE__ << ":<" << __func__ << ">:" << #x << "=" << x << endl;
#define debug(x) cerr << "line" << __LINE__ << ":<" << __func__ << ">:" << x << endl;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

using namespace std;
typedef long long ll;

template<typename T>
vector<vector<T>> vec2d(int n, int m, T v){
    return vector<vector<T>>(n, vector<T>(m, v));
}

template<typename T>
vector<vector<vector<T>>> vec3d(int n, int m, int k, T v){
    return vector<vector<vector<T>>>(n, vector<vector<T>>(m, vector<T>(k, v)));
}

template<typename T>
void print_vector(vector<T> v, char delimiter=' '){
    if(v.empty()) {
        cout << endl;
        return;
    }
    for(int i = 0; i+1 < v.size(); i++) cout << v[i] << delimiter;
    cout << v.back() << endl;
}

template<typename T>
vector<T> naive_convolution(vector<T> u, vector<T> v){
    int n = u.size(), m = v.size();
    vector<T> ans(n+m-1);
    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            ans[i+j] += u[i]*v[j];
        }
    }
    return ans;
}

/**
 * f(x) = g(x)*h(x) + r(x)となる(h(x), r(x))を求めます
 */ 
template<typename T>
pair<vector<T>, vector<T>> naive_divide(vector<T> f, vector<T> g){
    int n = f.size()-1;
    int m = g.size()-1;
    if(n < m) return make_pair(vector<T>(1, T(0)), f);
    int k = n-m;
    vector<T> h(k+1);
    T iv = T(1)/g[m];
    for(int i = k; i >= 0; i--){
        h[i] = f[i+m]*iv;
        for(int j = 0; j <= m; j++){
            f[i+j] -= g[j]*h[i];
        }
    }
    return make_pair(h, f);
}

using mint = atcoder::modint998244353;


/**
 * aで表される多項式にxを代入したときの値を返します。
 */ 
mint calc(vector<mint> a, mint x){
    mint cur = 1;
    mint ans = 0;
    for(auto b : a){
        ans += b*cur;
        cur *= x;
    }
    return ans;
}

/**
 * f(x_i) = y_iを満たす多項式を返します(O(N^2))
 * verified: https://atcoder.jp/contests/abc137/submissions/20140907
 */ 
vector<mint> lagrange_interpolation(vector<mint> x, vector<mint> y){
    assert(x.size() == y.size());
    int n = x.size();
    vector<vector<mint>> dp(2, vector<mint>(n+1));
    dp[0][0] = x[0]*(-1);
    dp[0][1] = 1;
    for(int i = 1; i < n; i++){
        int cur = i%2, prev = (i+1)%2;
        dp[cur][0] = dp[prev][0]*x[i]*(-1);
        for(int j = 1; j <= i+1; j++){
            dp[cur][j] = dp[prev][j-1]-dp[prev][j]*x[i];
        }
    }
    vector<mint> f(n+1);
    for(int i = 0; i <= n; i++) f[i] = dp[(n-1)%2][i];
    vector<mint> ans(n);
    for(int i = 0; i < n; i++){
        mint prod = 1;
        for(int j = 0; j < n; j++){
            if(i != j) prod *= (x[i]-x[j]);
        }
        // (x-x[i])*q = f
        vector<mint> q(n);
        q[n-1] = 1;
        for(int j = n-1; j >= 1; j--){
            q[j-1] = x[i]*q[j]+f[j];
        }
        mint coef = y[i]*prod.inv();
        for(int j = 0; j < n; j++) ans[j] += coef*q[j];
    }
    return ans;
}

ostream& operator<<(ostream& os, const mint& m){
    os << m.val();
    return os;
}


int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(10) << fixed;
    int n, X; cin >> n >> X;
    vector<mint> x(n), y(n);
    for(int i = 0; i < n; i++){
        int xx; cin >> xx; x[i] = xx;
        int yy; cin >> yy; y[i] = yy;
    }
    vector<mint> v = naive_convolution<mint>({-x[0], mint(1)}, {-x[1], mint(1)});
    for(int i = 2; i < n; i++){
        v = naive_convolution<mint>(v, {-x[i], mint(1)});
    }
    vector<mint> p(n);
    vector<mint> p_inv(n);
    for(int i = 0; i < n; i++){
        auto [pp, r] = naive_divide<mint>(v, {-x[i], mint(1)});
        p[i] = calc(pp, x[i]);
        p_inv[i] = p[i].inv();
    }
    vector<mint> sum(n);
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            if(i != j) {
                sum[j] += y[j];
                sum[i] -= p[i]*p_inv[j]*y[j];
            }
        }
    }
    auto ans = lagrange_interpolation(x, sum);
    cout << calc(ans, mint(X)) << endl;
}