Arkady has got an infinite plane painted in color 0. Then he draws n rectangles filled with paint with sides parallel to the Cartesian coordinate axes, one after another. The color of the i-th rectangle is i (rectangles are enumerated from 1 to n in the order he draws them). It is possible that new rectangles cover some of the previous ones completely or partially.

Count the number of different colors on the plane after Arkady draws all the rectangles.

Input

The first line contains a single integer n (1 ≤ n ≤ 100 000) — the number of rectangles.

The i-th of the next n lines contains 4 integers x_1, y_1, x_2 and y_2 (-10^9 ≤ x_1 < x_2 ≤ 10^9, -10^9 ≤ y_1 < y_2 ≤ 10^9) — the coordinates of corners of the i-th rectangle.

Output

In the single line print the number of different colors in the plane, including color 0.

Examples

Input

5
-1 -1 1 1
-4 0 0 4
0 0 4 4
-4 -4 0 0
0 -4 4 0


Output

5

Input

4
0 0 4 4
-4 -4 0 0
0 -4 4 0
-2 -4 2 4


Output

5

Note

<image> That's how the plane looks in the first sample

<image> That's how the plane looks in the second sample

0 = white, 1 = cyan, 2 = blue, 3 = purple, 4 = yellow, 5 = red. 
#include <bits/stdc++.h>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
inline int read() {
  int f = 1, x = 0;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1LL;
  } while (ch < '0' || ch > '9');
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return f * x;
}
template <class T>
inline void chmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void chmin(T &a, T b) {
  if (a > b) a = b;
}
inline void swap(int &a, int &b) {
  int c = a;
  a = b;
  b = c;
}
using namespace std;
const int N = 4e5 + 10;
set<int> seg[N << 2];
int mx[N << 2], mn[N << 2];
struct node {
  int x1, y1, x2, y2;
} a[N];
inline void show(node a) { printf("%d %d %d %d\n", a.x1, a.y1, a.x2, a.y2); }
vector<int> ins[N], era[N];
int n, px[N << 1], py[N << 1], totx, toty, lenx, leny, ok[N];
inline void upd(int cur) {
  int mxp;
  if (seg[cur].size()) {
    mxp = *(--seg[cur].end());
  } else {
    mxp = -1;
  }
  int mxt = max(mx[cur << 1], mx[cur << 1 | 1]);
  int mnt = min(mn[cur << 1], mn[cur << 1 | 1]);
  if (mxp > mxt) {
    if (ok[mxp] || mxp < mnt) {
      mx[cur] = -1;
    } else {
      mx[cur] = mxp;
    }
  } else {
    mx[cur] = mxt;
  }
  mn[cur] = max(mxp, mnt);
}
inline void cover(int cur, int L, int R, int l, int r, int id, int fl) {
  if (l > r) return;
  if (l <= L && r >= R) {
    if (fl == 1) {
      seg[cur].insert(id);
    } else if (fl == -1) {
      seg[cur].erase(id);
    }
    upd(cur);
    return;
  }
  int mid = (L + R) >> 1;
  if (l <= mid) cover(cur << 1, L, mid, l, r, id, fl);
  if (r > mid) cover(cur << 1 | 1, mid + 1, R, l, r, id, fl);
  upd(cur);
}
inline void norx(int &x) {
  x = (int)(lower_bound(px + 1, px + lenx + 1, x) - px);
}
inline void nory(int &x) {
  x = (int)(lower_bound(py + 1, py + leny + 1, x) - py);
}
int main() {
  memset(mx, -1, sizeof(mx));
  n = read();
  for (int i = (1); i <= (n); ++i)
    a[i].x1 = read(), a[i].y1 = read(), a[i].x2 = read(), a[i].y2 = read();
  for (int i = (1); i <= (n); ++i) {
    px[++totx] = a[i].x1;
    px[++totx] = a[i].x2;
    py[++toty] = a[i].y1;
    py[++toty] = a[i].y2;
  }
  sort(px + 1, px + totx + 1);
  sort(py + 1, py + toty + 1);
  lenx = (int)(unique(px + 1, px + totx + 1) - px - 1);
  leny = (int)(unique(py + 1, py + toty + 1) - py - 1);
  for (int i = (1); i <= (n); ++i) {
    norx(a[i].x1);
    norx(a[i].x2);
    nory(a[i].y1);
    nory(a[i].y2);
  }
  for (int i = (1); i <= (n); ++i) {
    ins[a[i].x1].push_back(i);
    era[a[i].x2].push_back(i);
  }
  for (int t = (1); t <= (lenx); ++t) {
    for (auto cur : ins[t]) {
      cover(1, 1, leny, a[cur].y1, a[cur].y2 - 1, cur, 1);
    }
    for (auto cur : era[t]) {
      cover(1, 1, leny, a[cur].y1, a[cur].y2 - 1, cur, -1);
    }
    while (~mx[1]) {
      ok[mx[1]] = 1;
      cover(1, 1, leny, a[mx[1]].y1, a[mx[1]].y2 - 1, mx[1], 0);
    }
  }
  int ans = 1;
  for (int i = (1); i <= (n); ++i)
    if (ok[i]) ++ans;
  printf("%d\n", ans);
  return 0;
}
