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;
struct node {
  int l, r, x;
  long long sum, lazy;
  node *lt, *rt;
  node(int a, int b) : l(a), r(b), x(-1), sum(0), lt(NULL), rt(NULL), lazy(0) {
    if (r == l)
      x = l + 1;
    else {
      int mid = (l + r) >> 1;
      lt = new node(l, mid);
      rt = new node(mid + 1, r);
    }
  }
  ~node() {
    if (lt) delete lt, delete rt;
  }
  void pass() {
    lt->x = rt->x = x;
    lt->sum += lazy * (lt->r - lt->l + 1);
    rt->sum += lazy * (rt->r - rt->l + 1);
    lt->lazy += lazy;
    rt->lazy += lazy;
    lazy = 0;
  }
  void update(int y, int a, int b) {
    if (a <= l && b >= r && x >= 0) {
      sum += (long long)(r - l + 1) * abs(y - x);
      lazy += abs(y - x);
      x = y;
    } else {
      if (x >= 0 && lazy) pass();
      int mid = (l + r) >> 1;
      if (a <= mid) lt->update(y, a, b);
      if (b > mid) rt->update(y, a, b);
      sum = lt->sum + rt->sum;
      x = -1;
      if (a <= l && b >= r) x = y;
    }
  }
  long long query(int a, int b) {
    if (a <= l && b >= r) return sum;
    if (x >= 0) pass();
    long long res = 0;
    int mid = (l + r) >> 1;
    if (a <= mid) res += lt->query(a, b);
    if (b > mid) res += rt->query(a, b);
    return res;
  }
};
int main() {
  int n, m;
  cin >> n >> m;
  node t(0, n - 1);
  for (int i = 0; i < m; i++) {
    int op;
    cin >> op;
    if (op == 1) {
      int l, r, x;
      cin >> l >> r >> x;
      t.update(x, l - 1, r - 1);
    } else {
      int l, r;
      cin >> l >> r;
      cout << t.query(l - 1, r - 1) << endl;
      ;
    }
  }
  return 0;
}
