#include <bits/stdc++.h>
#pragma GCC optimize(2)
#define ld long double
#define int long long
using namespace std;
int l[100005], r[100005], sum[100005], n, ssum[100005];
const int inf = 1000000000000000000;
struct XD {
	int k, b;
};
struct node {
	int l, r;
	XD maxn, minn;
} b[400005];
inline void build(int p, int l, int r) {
	b[p].l = l, b[p].r = r;
	b[p].minn = XD{0, inf}, b[p].maxn = XD{0, -inf};
	if (l == r) return;
	build(2 * p, l, (l + r) >> 1);
	build(2 * p + 1, ((l + r) >> 1) + 1, r);
}
inline int calc(XD p, int q) {
	return p.k * q + p.b;
}
inline void changemax(int p, XD op) {
	int mid = (b[p].l + b[p].r) >> 1;
	if (calc(op, ssum[mid]) > calc(b[p].maxn, ssum[mid])) {
		swap(op, b[p].maxn);
	}
	if (b[p].l == b[p].r) return;
	if (op.k < b[p].maxn.k) changemax(2 * p, op);
	else if (op.k > b[p].maxn.k) changemax(2 * p + 1, op);
}
inline void changemin(int p, XD op) {
	int mid = (b[p].l + b[p].r) >> 1;
	if (calc(op, ssum[mid]) < calc(b[p].minn, ssum[mid])) {
		swap(op, b[p].minn);
	}
	if (b[p].l == b[p].r) return;
	if (op.k < b[p].minn.k) changemin(2 * p + 1, op);
	else if (op.k > b[p].minn.k) changemin(2 * p, op);
}
inline int querymax(int p, int q) {
	if (b[p].l == b[p].r) {
		return calc(b[p].maxn, ssum[q]);
	}
	if (q <= (b[p].l + b[p].r) / 2) {
		return max(calc(b[p].maxn, ssum[q]), querymax(2 * p, q));
	}
	return max(calc(b[p].maxn, ssum[q]), querymax(2 * p + 1, q));
}
inline int querymin(int p, int q) {
	if (b[p].l == b[p].r) {
		return calc(b[p].minn, ssum[q]);
	}
	if (q <= (b[p].l + b[p].r) / 2) {
		return min(calc(b[p].minn, ssum[q]), querymin(2 * p, q));
	}
	return min(calc(b[p].minn, ssum[q]), querymin(2 * p + 1, q));
}
signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> l[i] >> r[i];
	}
	for (int i = 1; 2 * i <= n; i++) {
		swap(l[i], l[n - i + 1]);
		swap(r[i], r[n - i + 1]);
	}
	for (int i = 1; i <= n; i++) {
		l[i] <<= 1, r[i] <<= 1;
		sum[i] = sum[i - 1] + (r[i] - l[i]) / 2 * (r[i] + l[i]);
		ssum[i] = ssum[i - 1] + (r[i] - l[i]);
	}
	build(1, 1, n);
	int ans = -1;
	int lst = 0;
	for (int i = 1; i < n; i++) {
//		cout << querymax(1, i) << " " << querymin(1, i) << " " << ssum[i] << " " << sum[i] << " " << l[i] << " " << r[i] << "\n";
		if (querymax(1, i) <= sum[i] && querymin(1, i) >= sum[i]) {
			ans = max(ans, i - lst);
			lst = i;
		}
		changemax(1, XD{l[i], -ssum[i]*l[i] + sum[i]});
		changemin(1, XD{r[i], -ssum[i]*r[i] + sum[i]});
//		if (n - i == 3) {
//			for (int j = 1; j < i; j++) {
//				if (sum[i] - sum[j] < l[j] * (ssum[i] - ssum[j]) || sum[i] - sum[j] > r[j] * (ssum[i] - ssum[j])) {
//					cout << j << "[]\n";
//				}
//			}
//		}
//		cout << b[1].minn.k << " " << b[1].minn.b << " " << b[1].maxn.k << ' ' << b[1].maxn.b << " " << calc(b[1].minn, ssum[i + 1]) << " " << calc(b[1].maxn, ssum[i + 1]) << "[[]%$&^\n";
	}
	ans = max(ans, n - lst);
	cout << ans;
}
//A 题不会做 Ai=Bi 的情况，快写完了才想起来，我炸了

//尼玛这题也读错了一次，读错小子
//最开始还以为只要捞一叠出来就行，没想到原来是这个意思