Problem: This year Alex has finished school, and now he is a first-year student of Berland State University. For him it was a total surprise that even though he studies programming, he still has to attend physical education lessons. The end of the term is very soon, but, unfortunately, Alex still hasn't attended a single lesson!

Since Alex doesn't want to get expelled, he wants to know the number of working days left until the end of the term, so he can attend physical education lessons during these days. But in BSU calculating the number of working days is a complicated matter:

There are n days left before the end of the term (numbered from 1 to n), and initially all of them are working days. Then the university staff sequentially publishes q orders, one after another. Each order is characterised by three numbers l, r and k:

  * If k = 1, then all days from l to r (inclusive) become non-working days. If some of these days are made working days by some previous order, then these days still become non-working days; 
  * If k = 2, then all days from l to r (inclusive) become working days. If some of these days are made non-working days by some previous order, then these days still become working days. 



Help Alex to determine the number of working days left after each order!

Input

The first line contains one integer n, and the second line — one integer q (1 ≤ n ≤ 109, 1 ≤ q ≤ 3·105) — the number of days left before the end of the term, and the number of orders, respectively.

Then q lines follow, i-th line containing three integers li, ri and ki representing i-th order (1 ≤ li ≤ ri ≤ n, 1 ≤ ki ≤ 2).

Output

Print q integers. i-th of them must be equal to the number of working days left until the end of the term after the first i orders are published.

Example

Input

4
6
1 2 1
3 4 1
2 3 2
1 3 2
2 4 1
1 4 2


Output

2
0
2
3
1
4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
namespace IOS {
template <class T>
inline void read(T& x) {
  register char c = getchar(), d = 0;
  while (c < '0' || c > '9') d = c, c = getchar();
  x = 0;
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
  if (d == '-') x = -x;
}
template <class T>
inline void wt(T x) {
  if (x / 10) wt(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline void wtsp(T x) {
  if (x < 0) putchar('-'), x = -x;
  wt(x), putchar(' ');
}
template <class T>
inline void wtln(T x) {
  if (x < 0) putchar('-'), x = -x;
  wt(x), putchar('\n');
}
}  // namespace IOS
using namespace IOS;
int n, m;
int op, l, r;
int ans;
namespace OldDriverTree {
struct SEG {
  int lpos;
  mutable int rpos;
  mutable int val;
  inline int len() const { return rpos - lpos + 1; }
  inline void mov(int dlt) { lpos += dlt, rpos += dlt; }
  inline void rev(int sum) {
    lpos = sum - lpos, rpos = sum - rpos;
    std::swap(lpos, rpos);
  }
  inline SEG(int LPOS, int RPOS = -1, int VAL = -1)
      : lpos(LPOS), rpos(RPOS), val(VAL){};
  inline bool operator<(const SEG& seg) const { return lpos < seg.lpos; }
};
class ODT {
 private:
  std::set<SEG> data;
  std::set<SEG>::iterator iL, iR;
  inline std::set<SEG>::iterator split(int p) {
    std::set<SEG>::iterator it = data.lower_bound(p);
    if (it->lpos == p) return it;
    --it;
    SEG np = *it;
    it->rpos = p - 1, np.lpos = p;
    return data.insert(np).first;
  }
  inline std::set<SEG>::iterator insert(const SEG& s) {
    std::set<SEG>::iterator it = data.insert(s).first;
    if (it != data.begin()) {
      std::set<SEG>::iterator iL = it;
      --iL;
      if (iL->rpos + 1 == it->lpos && iL->val == it->val) {
        iL->rpos = it->rpos;
        data.erase(it), it = iL;
      }
    }
    std::set<SEG>::iterator iR = it;
    ++iR;
    if (iR != data.end()) {
      if (it->rpos + 1 == iR->lpos && iR->val == it->val) {
        it->rpos = iR->rpos;
        data.erase(iR);
      }
    }
    return it;
  }
  inline void erase() { data.erase(iL, iR); }

 public:
  inline void Ini(int n) {
    data.clear();
    insert(SEG(1, n, 1));
    insert(SEG(n + 1));
  }
  inline void Eql(int l, int r, int val) {
    iL = split(l), iR = split(r + 1);
    for (std::set<SEG>::iterator i = iL; i != iR; ++i) ans -= i->val * i->len();
    erase();
    insert(SEG(l, r, val));
    ans += val * (r - l + 1);
  }
  inline void Prt() {
    for (SEG it : data)
      for (int i = it.lpos; i <= it.rpos; ++i) wtsp(it.val);
    putchar('\n');
  }
};
}  // namespace OldDriverTree
OldDriverTree::ODT T;
int main() {
  read(n), read(m);
  T.Ini(n), ans = n;
  for (int i = 1; i <= m; ++i) {
    read(l), read(r), read(op);
    T.Eql(l, r, --op);
    wtln(ans);
  }
}