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 int N = 1e5 + 3;
long long seg[1 << 18];
long long lazy[1 << 18];
inline int ab(int x) { return x < 0 ? -x : x; }
void ust(int l, int r, int a, int b, int x, int i) {
  if (lazy[i] > 0) {
    seg[i] += (r - l + 1) * lazy[i];
    if (l != r) {
      lazy[i + i + 1] += lazy[i];
      lazy[i + i + 2] += lazy[i];
    }
    lazy[i] = 0;
  }
  if (l > b || r < a) return;
  if (l >= a && r <= b) {
    seg[i] += 1LL * (r - l + 1) * x;
    if (l != r) {
      lazy[i + i + 1] += x;
      lazy[i + i + 2] += x;
    }
    return;
  }
  int mid = l + r >> 1;
  ust(l, mid, a, b, x, i + i + 1);
  ust(mid + 1, r, a, b, x, i + i + 2);
  seg[i] = seg[i + i + 1] + seg[i + i + 2];
}
long long qst(int l, int r, int a, int b, int i) {
  if (lazy[i] > 0) {
    seg[i] += (r - l + 1) * lazy[i];
    if (l != r) {
      lazy[i + i + 1] += lazy[i];
      lazy[i + i + 2] += lazy[i];
    }
    lazy[i] = 0;
  }
  if (l > b || r < a) return 0;
  if (l >= a && r <= b) return seg[i];
  int mid = l + r >> 1;
  return qst(l, mid, a, b, i + i + 1) + qst(mid + 1, r, a, b, i + i + 2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  set<pair<pair<int, int>, int> > my, my2;
  set<pair<pair<int, int>, int> >::iterator it, it2;
  for (int i = 0; i < n; ++i) {
    my.insert(make_pair(make_pair(i, i), i + 1));
    my2.insert(make_pair(make_pair(-i, -i), i + 1));
  }
  for (int qq = 0; qq < m; ++qq) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r, x;
      cin >> l >> r >> x;
      --l, --r;
      it2 = my2.upper_bound(make_pair(make_pair(-l, 1), 0));
      if (it2 != my2.end()) {
        auto elem = *it2;
        auto elem2 = elem;
        elem.first.first *= -1;
        elem.first.second *= -1;
        if (elem.first.second >= l) {
          ust(0, n - 1, l, min(r, elem.first.second), ab(elem.second - x), 0);
          my.erase(elem);
          my2.erase(elem2);
          my.insert(make_pair(make_pair(elem.first.first, l - 1), elem.second));
          my2.insert(
              make_pair(make_pair(-elem.first.first, -l + 1), elem.second));
          if (elem.first.second > r) {
            my.insert(
                make_pair(make_pair(r + 1, elem.first.second), elem.second));
            my2.insert(
                make_pair(make_pair(-r - 1, -elem.first.second), elem.second));
          }
        }
      }
      for (;;) {
        it = my.lower_bound(make_pair(make_pair(l, -1), -1));
        if (it == my.end()) break;
        auto elem = *it;
        if (elem.first.first > r) break;
        if (elem.first.second <= r) {
          ust(0, n - 1, elem.first.first, elem.first.second,
              ab(elem.second - x), 0);
          my.erase(elem);
          my2.erase(make_pair(make_pair(-elem.first.first, -elem.first.second),
                              elem.second));
        } else {
          my.erase(elem);
          my2.erase(make_pair(make_pair(-elem.first.first, -elem.first.second),
                              elem.second));
          ust(0, n - 1, elem.first.first, r, ab(elem.second - x), 0);
          my.insert(
              make_pair(make_pair(r + 1, elem.first.second), elem.second));
          my2.insert(
              make_pair(make_pair(-r - 1, -elem.first.second), elem.second));
        }
      }
      my.insert(make_pair(make_pair(l, r), x));
      my2.insert(make_pair(make_pair(-l, -r), x));
    } else {
      int l, r;
      cin >> l >> r;
      --l, --r;
      cout << qst(0, n - 1, l, r, 0) << '\n';
    }
  }
}
