Problem:
Score : 1000 points

Problem Statement

Snuke arranged N colorful balls in a row.
The i-th ball from the left has color c_i and weight w_i.

He can rearrange the balls by performing the following two operations any number of times, in any order:

Operation 1: Select two balls with the same color. If the total weight of these balls is at most X, swap the positions of these balls.

Operation 2: Select two balls with different colors. If the total weight of these balls is at most Y, swap the positions of these balls.

How many different sequences of colors of balls can be obtained? Find the count modulo 10^9 + 7.

Constraints

1 ≤ N ≤ 2 × 10^5

1 ≤ X, Y ≤ 10^9

1 ≤ c_i ≤ N

1 ≤ w_i ≤ 10^9

X, Y, c_i, w_i are all integers.

Input

Input is given from Standard Input in the following format:

N X Y
c_1 w_1
:
c_N w_N

Output

Print the answer.

Sample Input 1

4 7 3
3 2
4 3
2 1
4 4

Sample Output 1

2

The sequence of colors (2,4,3,4) can be obtained by swapping the positions of the first and third balls by operation 2.

It is also possible to swap the positions of the second and fourth balls by operation 1, but it does not affect the sequence of colors.

Sample Input 2

1 1 1
1 1

Sample Output 2

1

Sample Input 3

21 77 68
16 73
16 99
19 66
2 87
2 16
7 17
10 36
10 68
2 38
10 74
13 55
21 21
3 7
12 41
13 88
18 6
2 12
13 87
1 9
2 27
13 15

Sample Output 3

129729600

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

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

#define ms(s, n) memset(s, n, sizeof(s))
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define FORd(i, a, b) for (int i = (a) - 1; i >= (b); i--)
#define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define sz(a) int((a).size())
#define present(t, x) (t.find(x) != t.end())
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vii;
const int MOD = (int) 1e9 + 7;
const int INF = (int) 1e9;
const ll LINF = (ll) 1e18;
const ld PI = acos((ld) -1);
const ld EPS = 1e-9;
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;}
template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;}
inline ll isqrt(ll k) {ll r = sqrt(k) + 1; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);}
inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}
#define db(x) cout << #x << " = " << (x) << "\n";

const int maxn = 2e5 + 5;
int n, x, y;
int c[maxn];
int w[maxn];
int dj[maxn];
int sz[maxn];
vi g[maxn];
int fac[maxn];
int ifac[maxn];
map<int, int> hs[maxn];

int binom(int a, int b) {
	if (a < 0 || a > b) return 0;
	if (!a || a == b) return 1;
	return mult(fac[b], mult(ifac[a], ifac[b - a]));
}

void init() {
	FOR(i, 0, n) dj[i] = i, sz[i] = 1;
}
int find(int u) {
	return dj[u] == u ? dj[u] : dj[u] = find(dj[u]);
}
void join(int u, int v) {
	int p = find(u);
	int q = find(v);
	if (p != q) {
		dj[p] = q;
		sz[q] += sz[p];
	}
}

int compare(int i, int j) {
	if (w[i] != w[j]) return w[i] < w[j];
	return i < j;
}

void solve() {
	fac[0] = 1; FOR(i, 1, maxn) fac[i] = mult(fac[i - 1], i);
	FOR(i, 0, maxn) ifac[i] = inv(fac[i]);
	cin >> n >> x >> y;
	FOR(i, 0, n) cin >> c[i] >> w[i];
	vi ver; FOR(i, 0, n) ver.pb(i);
	sort(all(ver), compare);
	int u = ver[0], v = -1, t = -1;
	FOR(i, 1, n) if (ver[i] != u) {
		if (v == -1) {
			v = ver[i];
		}
		else if (t == -1) {
			t = ver[i];
			break;
		}
	}
	init();
	FOR(i, 0, n) {
		if (c[u] == c[i] && w[u] + w[i] <= x) {
			join(u, i);
		}
		else if (c[u] != c[i] && w[u] + w[i] <= y) {
			join(u, i);
		}
	}
	if (v != -1) {
		FOR(i, 0, n) {
			if (c[v] == c[i] && w[v] + w[i] <= x) {
				join(v, i);
			}
			else if (c[v] != c[i] && w[v] + w[i] <= y) {
				join(v, i);
			}
		}
	}
	v = t;
	if (v != -1) {
		FOR(i, 0, n) {
			if (c[v] == c[i] && w[v] + w[i] <= x) {
				join(v, i);
			}
			else if (c[v] != c[i] && w[v] + w[i] <= y) {
				join(v, i);
			}
		}
	}
	FOR(i, 0, n) {
		hs[find(i)][c[i]]++;
	}
	int ans = 1;
	FOR(i, 0, n) {
		int cnt = 0;
		FORall(it, hs[i]) {
			cnt += it->se;
			ans = mult(ans, binom(it->se, cnt));
		}
	}
	cout << ans << "\n";
}

int main() {
	ios_base::sync_with_stdio(0); //cin.tie(0);
#ifdef _LOCAL_
	freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout);
#endif
	solve();
#ifdef _LOCAL_
	//printf("\nTime elapsed: %dms", 1000 * clock() / CLOCKS_PER_SEC);
#endif
	return 0;
}