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 Tree {
  Tree *ls, *rs;
  int l, r;
  long long sum;
  bool same;
  int x;
  long long add;
  Tree() {}
  ~Tree() {
    if (ls) delete ls;
    if (rs) delete rs;
  }
  void up() {
    if (ls->same && rs->same && ls->x == rs->x) {
      same = 1;
      x = ls->x;
      add = 0;
    } else
      same = 0, add = 0;
    sum = rs->sum + ls->sum;
  }
  void down() {
    if (add) {
      ls->add += add;
      ls->sum += add * (ls->r - ls->l);
      ls->x = x;
      rs->add += add;
      rs->sum += add * (rs->r - rs->l);
      rs->x = x;
      add = 0;
    }
  }
  Tree(int l, int r) : l(l), r(r) {
    ls = rs = NULL;
    sum = 0;
    same = 0;
    x = l;
    add = 0;
    if (r - l <= 1) {
      same = 1;
      return;
    }
    int m = (l + r) >> 1;
    ls = new Tree(l, m);
    rs = new Tree(m, r);
    up();
  }
  void gao(int L, int R, int X) {
    if (L <= l && r <= R) {
      if (same) {
        long long v = 1LL * abs(X - x);
        x = X;
        add += v;
        sum += v * (r - l);
        return;
      }
    }
    down();
    int m = (l + r) >> 1;
    if (L < m) ls->gao(L, R, X);
    if (R > m) rs->gao(L, R, X);
    up();
  }
  long long ask(int L, int R) {
    if (L <= l && r <= R) return sum;
    int m = (l + r) >> 1;
    long long ans = 0;
    down();
    if (L < m) ans += ls->ask(L, R);
    if (R > m) ans += rs->ask(L, R);
    up();
    return ans;
  }
};
int n, m;
int main() {
  ios_base::sync_with_stdio(false);
  while (cin >> n >> m) {
    Tree* t = new Tree(1, n + 1);
    int ty, a, b, c;
    for (int i = 0; i < m; i++) {
      cin >> ty;
      if (ty == 1) {
        cin >> a >> b >> c;
        t->gao(a, b + 1, c);
      } else {
        cin >> a >> b;
        cout << t->ask(a, b + 1) << endl;
      }
    }
    delete t;
  }
  return 0;
}
