Problem: There is a square grid of size n × n. Some cells are colored in black, all others are colored in white. In one operation you can select some rectangle and color all its cells in white. It costs min(h, w) to color a rectangle of size h × w. You are to make all cells white for minimum total cost.

The square is large, so we give it to you in a compressed way. The set of black cells is the union of m rectangles.

Input

The first line contains two integers n and m (1 ≤ n ≤ 10^{9}, 0 ≤ m ≤ 50) — the size of the square grid and the number of black rectangles.

Each of the next m lines contains 4 integers x_{i1} y_{i1} x_{i2} y_{i2} (1 ≤ x_{i1} ≤ x_{i2} ≤ n, 1 ≤ y_{i1} ≤ y_{i2} ≤ n) — the coordinates of the bottom-left and the top-right corner cells of the i-th black rectangle.

The rectangles may intersect.

Output

Print a single integer — the minimum total cost of painting the whole square in white.

Examples

Input


10 2
4 1 5 10
1 4 10 5


Output


4


Input


7 6
2 1 2 1
4 2 4 3
2 5 2 5
2 3 5 3
1 2 1 2
3 2 5 3


Output


3

Note

The examples and some of optimal solutions are shown on the pictures below.

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N, M, Aux[105], source, sink;
vector<pair<int, int> > G[405];
int Rev[805 * 805], Cap[805 * 805], F[805 * 805];
pair<int, int> L[205], R[205];
int cntL, cntR;
const int INF = 1000000005;
int Ind[405], Level[405];
bool Used[405];
queue<int> Q;
pair<pair<int, int>, pair<int, int> > Rectangle[55];
int X[105], Y[105], cntX, cntY, cntE;
void Read() {
  scanf("%d%d", &N, &M);
  for (int i = 1; i <= M; i++) {
    int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    Rectangle[i] = make_pair(make_pair(x1, y1), make_pair(x2, y2));
    X[++cntX] = x1;
    X[++cntX] = x2;
    Y[++cntY] = y1;
    Y[++cntY] = y2;
  }
  sort(X + 1, X + cntX + 1);
  sort(Y + 1, Y + cntY + 1);
}
void eraseDuplicates(int X[], int& cntX) {
  int cnt = 0;
  for (int i = 1; i <= cntX; i++) {
    if (i == 1 || X[i] != X[i - 1]) Aux[++cnt] = X[i];
  }
  cntX = cnt;
  for (int i = 1; i <= cntX; i++) X[i] = Aux[i];
}
void addEdge(int x, int y, int cap, int revCap) {
  ++cntE;
  G[x].push_back(make_pair(y, cntE));
  Cap[cntE] = cap;
  Rev[cntE] = cntE + 1;
  ++cntE;
  G[y].push_back(make_pair(x, cntE));
  Cap[cntE] = revCap;
  Rev[cntE] = cntE - 1;
}
void buildRange(int X[], int cntX, pair<int, int> L[], int& cntL) {
  for (int i = 1; i <= cntX; i++) {
    L[++cntL] = make_pair(X[i], X[i]);
    if (i < cntX && X[i] + 1 < X[i + 1]) {
      L[++cntL] = make_pair(X[i] + 1, X[i + 1] - 1);
    }
  }
}
void computeLimits(int x1, int y1, int x2, int y2, int& leftL, int& rightL,
                   int& leftR, int& rightR) {
  for (int j = 1; j <= cntL; j++) {
    if (L[j].first >= x1 && L[j].second <= x2) {
      leftL = j;
      break;
    }
  }
  for (int j = 1; j <= cntL; j++)
    if (L[j].first >= x1 && L[j].second <= x2) {
      rightL = j;
    }
  for (int j = 1; j <= cntR; j++) {
    if (R[j].first >= y1 && R[j].second <= y2) {
      leftR = j;
      break;
    }
  }
  for (int j = 1; j <= cntR; j++)
    if (R[j].first >= y1 && R[j].second <= y2) {
      rightR = j;
    }
}
void buildNetwork() {
  source = 0;
  sink = cntL + cntR + 1;
  for (int i = 1; i <= cntL; i++) {
    addEdge(source, i, L[i].second - L[i].first + 1, 0);
  }
  for (int i = 1; i <= cntR; i++) {
    addEdge(i + cntL, sink, R[i].second - R[i].first + 1, 0);
  }
  for (int i = 1; i <= M; i++) {
    int x1 = Rectangle[i].first.first, y1 = Rectangle[i].first.second;
    int x2 = Rectangle[i].second.first, y2 = Rectangle[i].second.second;
    int leftL, rightL, leftR, rightR;
    computeLimits(x1, y1, x2, y2, leftL, rightL, leftR, rightR);
    for (int j = leftL; j <= rightL; j++)
      for (int k = leftR; k <= rightR; k++)
        addEdge(j, k + cntL,
                min(L[j].second - L[j].first + 1, R[k].second - R[k].first + 1),
                0);
  }
}
bool BFS() {
  for (int i = source; i <= sink; i++) Used[i] = false, Level[i] = -1;
  Level[source] = 0;
  Used[source] = true;
  Q.push(source);
  while (!Q.empty()) {
    int node = Q.front();
    Q.pop();
    if (node == sink) continue;
    for (int j = 0; j < G[node].size(); j++) {
      int neighb = G[node][j].first, edge = G[node][j].second;
      if (!Used[neighb] && Cap[edge] - F[edge] > 0) {
        Used[neighb] = true;
        Q.push(neighb);
        Level[neighb] = Level[node] + 1;
      }
    }
  }
  return Used[sink];
}
int pushFlow(int node, int flow) {
  if (node == sink) return flow;
  for (; Ind[node] < G[node].size(); Ind[node]++) {
    int neighb = G[node][Ind[node]].first, edge = G[node][Ind[node]].second;
    if (Cap[edge] - F[edge] > 0 && Level[neighb] == Level[node] + 1) {
      int f = pushFlow(neighb, min(flow, Cap[edge] - F[edge]));
      if (f > 0) {
        F[edge] += f;
        F[Rev[edge]] -= f;
        return f;
      }
    }
  }
  return 0;
}
void init() {
  for (int i = source; i <= sink; i++) Ind[i] = 0;
}
void Dinic() {
  int maxFlow = 0;
  while (BFS()) {
    init();
    int flow = pushFlow(source, INF);
    while (flow > 0) {
      maxFlow += flow;
      flow = pushFlow(source, INF);
    }
  }
  printf("%d\n", maxFlow);
}
int main() {
  Read();
  eraseDuplicates(X, cntX);
  eraseDuplicates(Y, cntY);
  buildRange(X, cntX, L, cntL);
  buildRange(Y, cntY, R, cntR);
  buildNetwork();
  Dinic();
  return 0;
}