Problem: The Alice's computer is broken, so she can't play her favorite card game now. To help Alice, Bob wants to answer n her questions. 

Initially, Bob holds one card with number 0 in the left hand and one in the right hand. In the i-th question, Alice asks Bob to replace a card in the left or right hand with a card with number k_i (Bob chooses which of two cards he changes, Bob must replace exactly one card).

After this action, Alice wants the numbers on the left and right cards to belong to given segments (segments for left and right cards can be different). Formally, let the number on the left card be x, and on the right card be y. Then after the i-th swap the following conditions must be satisfied: a_{l, i} ≤ x ≤ b_{l, i}, and a_{r, i} ≤ y ≤ b_{r,i}.

Please determine if Bob can answer all requests. If it is possible, find a way to do it.

Input

The first line contains two integers n and m (2 ≤ n ≤ 100 000, 2 ≤ m ≤ 10^9) — the number of questions and the maximum possible value on the card.

Then n queries are described. Every description contains 3 lines.

The first line of the description of the i-th query contains a single integer k_i (0 ≤ k_i ≤ m) — the number on a new card.

The second line of the description of the i-th query contains two integers a_{l, i} and b_{l, i} (0 ≤ a_{l, i} ≤ b_{l, i} ≤ m) — the minimum and maximum values of the card at the left hand after the replacement.

The third line of the description of the i-th query contains two integers a_{r, i} and b_{r,i} (0 ≤ a_{r, i} ≤ b_{r,i} ≤ m) — the minimum and maximum values of the card at the right hand after the replacement.

Output

At the first line, print "Yes", if Bob can answer all queries, and "No" otherwise.

If Bob can answer all n queries, then at the second line print n numbers: a way to satisfy all requirements. If in i-th query Bob needs to replace the card in the left hand, print 0, otherwise print 1. If there are multiple answers, print any.

Examples

Input


2 10
3
0 3
0 2
2
0 4
0 2


Output


Yes
0 1 

Input


2 10
3
0 3
0 2
2
3 4
0 1


Output


No

Input


5 10
3
0 3
0 3
7
4 7
1 3
2
2 3
3 7
8
1 8
1 8
6
1 6
7 10


Output


Yes
1 0 0 1 0 

Solution (in CPP):

Solution:
// author: erray
#include <bits/stdc++.h>
 
using namespace std;
 template<typename A, typename B> string to_string(pair<A, B> p);
template<typename A, typename B, typename C> string to_string(tuple<A, B, C> t);
template<typename A, typename B, typename C, typename D> string to_string(tuple<A, B, C, D> t);

string to_string(string s) {
  return '"' + s + '"';
}

string to_string(char c) {
  return string("'") + c + "'";
}

string to_string(const char* c) {
  return to_string(string(c));
}

string to_string(bool b) {
  return (b ? "true" : "false");
}

template<size_t T> string to_string(bitset<T> bs) {
  return bs.to_string();
}

string to_string(vector<bool> v) {
  string res = "{";
  for (int i = 0; i < int(v.size()); ++i) {
    if (i > 0) {
      res += ", ";
    }
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}

template<typename T> string to_string(T v) {
  string res = "{";
  for (auto& e : v) {
    if (int(res.size()) > 1) {
      res += ", ";
    }
    res += to_string(e);
  }
  res += "}";
  return res;
}

template<typename A, typename B> string to_string(pair<A, B> p) {
  return '(' + to_string(p.first) + ", " + to_string(p.second) + ')';
}
template<typename A, typename B, typename C> string to_string(tuple<A, B, C> t) {
  return '(' + to_string(get<0>(t)) + ", " + to_string(get<1>(t)) + ", " + to_string(get<2>(t)) + '}';
}
template<typename A, typename B, typename C, typename D> string to_string(tuple<A, B, C, D> t) {
  return '(' + to_string(get<0>(t)) + ", " + to_string(get<1>(t)) + ", " + to_string(get<2>(t)) + ", " + to_string(get<3>(t)) + '}';   
}

void debug_out() {
  cerr << endl;
}

template<typename Head, typename... Tail> void debug_out(Head H, Tail... T) {
  cerr << "  " << to_string(H);
  debug_out(T...);
}

#ifdef DEBUG 
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) void(37)
#endif

const int inf = int(2e9);
struct SegTree {
  struct node {
    int tag = inf;
    int mx = -1;
    void modify(int x) {
      tag = x;
      mx = x;
    }
  };

  int n;
  vector<node> tree;

  node unite(const node& x, const node& y) {
    node res;
    res.mx = max(x.mx, y.mx);
    return res;
  }

  void push(int v, int rv) {
    if (tree[v].tag != inf) {
      tree[v + 1].modify(tree[v].tag);
      tree[rv].modify(tree[v].tag);
    }
    tree[v].tag = inf;
  }
  
  /*==========================================================================*/

  void pull(int v, int rv) {
    tree[v] = unite(tree[v + 1], tree[rv]);
  }

  void modify(int v, int l, int r, int ll, int rr, int x) {
    if (l >= ll && rr >= r) {
      tree[v].modify(x);
      return;
    }
    int mid = (l + r) >> 1;
    int rv = v + ((mid - l + 1) << 1);
    push(v, rv);
    if (mid >= ll) {
      modify(v + 1, l, mid, ll, rr, x);
    }
    if (mid < rr) {
      modify(rv, mid + 1, r, ll, rr, x);      
    }
    pull(v, rv);
  };

  node get(int v, int l, int r, int ll, int rr) {
    if (l >= ll && rr >= r) {    
      return tree[v];
    }
    int mid = (l + r) >> 1;
    int rv = v + ((mid - l + 1) << 1);
    push(v, rv);
    node res;
    if (mid >= ll) {
      if (mid < rr) {
        res = unite(get(v + 1, l, mid, ll, rr), get(rv, mid + 1, r, ll, rr));
      } else {   
        res = get(v + 1, l, mid, ll, rr);
      }
    } else {
      res = get(rv, mid + 1, r, ll, rr);
    }
    return res;
  }

  void modify(int ll, int rr, int x) {
    assert(ll >= 0 && rr < n && ll <= rr);
    modify(0, 0, n - 1, ll, rr, x);
  }

  node get(int ll, int rr) {
    assert(ll >= 0 && rr < n && ll <= rr);  
    return get(0, 0, n - 1, ll, rr);
  }

  SegTree() { }

  SegTree(int _n) : n(_n) {
    tree.resize((n << 1) - 1);
  }
};

void my_assert(bool x) {
  if (!x) {
    cout << "No\n";
    exit(0);
  }
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> k(n);
  vector<array<int, 2>> a(n);  
  vector<array<int, 2>> b(n);

  for (int i = 0; i < n; ++i) {
    cin >> k[i];
    cin >> a[i][0] >> a[i][1];
    cin >> b[i][0] >> b[i][1];
  }

  vector<int> all = k;
  all.push_back(-1);
  all.push_back(0);
  all.push_back(m + 1);
  sort(all.begin(), all.end());
  all.erase(unique(all.begin(), all.end()), all.end());
  int size = int(all.size());
  for (int i = 0; i < n; ++i) {
    k[i] = int(lower_bound(all.begin(), all.end(), k[i]) - all.begin());
  }
  
  debug(all, k);
  array<vector<array<int, 2>>, 2> ranges = {a, b};
  for (int j = 0; j < 2; ++j) {
    for (auto&[l, r] : ranges[j]) {
      l = int(lower_bound(all.begin(), all.end(), l) - all.begin());
      r = int(lower_bound(all.begin(), all.end(), r + 1) - all.begin()) - 1;  
      debug(l, r);
      my_assert(l <= r);    
    }
  }

  debug(ranges);
  debug(k);
  vector<vector<bool>> dp(n, vector<bool>(2));
  vector<vector<bool>> can(n, vector<bool>(2));
  vector<vector<int>> go(n, vector<int>(2));
  vector<vector<int>> pref(n + 1, vector<int>(2));  
  SegTree emp(size);
  array<SegTree, 2> ks = {emp, emp};
  for (int j = 0; j < 2; ++j) {
    ks[j].modify(1, 1, 0);
  }

  for (int i = 0; i < n; ++i) {
    /*
    #ifdef DEBUG
    for (int j = 0; j < 2; ++j) {
      for (int ind = 0; ind < size; ++ind) {
        debug(j, ind, ks[j].get(ind, ind).mx);
      }
    }
    #endif
    */
    for (int me = 0; me < 2; ++me) {
      if (ranges[me][i][0] > k[i] || ranges[me][i][1] < k[i]) {
        continue;
      }
      can[i][me] = true;
      int ot = me ^ 1;
      int take = ks[ot].get(ranges[ot][i][0], ranges[ot][i][1]).mx;
      --take;
      debug(i, me, take, ranges[ot][i]);
      if (take == -2 || pref[i][me] - pref[take + 1][me] != (i - take - 1)) {
        continue;
      }
      dp[i][me] = true;
      go[i][me] = take;
    }

    for (int j = 0; j < 2; ++j) {
      pref[i + 1][j] = pref[i][j] + can[i][j];
      if (dp[i][j]) {
        ks[j].modify(k[i], k[i], i + 1);  
      }
      if (ranges[j][i][0] > 0) {
        ks[j].modify(0, ranges[j][i][0] - 1, -1);
      }
      if (ranges[j][i][1] < size - 1) {
        ks[j].modify(ranges[j][i][1] + 1, size - 1, -1);
      }
    }
  }
                        
  debug(dp);   
  debug(go);
  debug(can);

  my_assert(dp[n - 1][0] || dp[n - 1][1]);
  int take = dp[n - 1][1];
  int till = go[n - 1][take];
  vector<int> ans(n);
  for (int i = n - 1; i >= 0; --i) {
    if (till == i) {
      take ^= 1;
      till = go[i][take];
    }
    ans[i] = take;
  } 
  cout << "Yes\n";
  for (int i = 0; i < n; ++i) {
    cout << ans[i] << ' ';
  }

}