DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

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


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = INF * (long long)INF;
const int MAX = 100100;
struct Fen {
  long long t[MAX];
  long long get(int r) {
    long long res = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) res += t[r];
    return res;
  }
  void add(int pos, long long val) {
    for (; pos < MAX; pos |= pos + 1) t[pos] += val;
  }
};
struct RangeFen {
  Fen a;
  Fen b;
  long long get(int r) { return a.get(r) * (r + 1) + b.get(r); }
  long long get(int l, int r) { return get(r) - get(l - 1); }
  void add(int r, long long val) {
    a.add(0, val);
    a.add(r + 1, -val);
    b.add(r + 1, (r + 1) * val);
  }
  void add(int l, int r, long long val) {
    add(r, val);
    if (l) add(l - 1, -val);
  }
} F;
set<pair<pair<int, int>, int> > seg;
void paint(int l, int r, int x) {
  set<pair<pair<int, int>, int> >::iterator it = seg.lower_bound(make_pair(
                                                make_pair(l, INF), INF)),
                                            jt;
  --it;
  vector<pair<pair<int, int>, int> > toAdd;
  while (it != seg.end() && it->first.first <= r) {
    if (it->first.first < l)
      toAdd.push_back(make_pair(make_pair(it->first.first, l - 1), it->second));
    if (it->first.second > r)
      toAdd.push_back(
          make_pair(make_pair(r + 1, it->first.second), it->second));
    int ll = max(l, it->first.first);
    int rr = min(r, it->first.second);
    int d = abs(x - it->second);
    F.add(ll, rr, d);
    jt = it;
    ++it;
    seg.erase(jt);
  }
  for (int i = (0); i < ((int)toAdd.size()); i++) {
    seg.insert(toAdd[i]);
  }
  seg.insert(make_pair(make_pair(l, r), x));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = (0); i < (n); i++) {
    seg.insert(make_pair(make_pair(i, i), i));
  }
  for (int i = (0); i < (m); i++) {
    int t, l, r;
    cin >> t >> l >> r;
    --l;
    --r;
    if (t == 1) {
      int x;
      cin >> x;
      --x;
      paint(l, r, x);
    } else {
      long long ans = F.get(l, r);
      cout << ans << "\n";
    }
  }
}
