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 maxn = 100010;
int n, m;
struct IntervalTree {
  long long sum[maxn << 3], setv[maxn << 3], val[maxn << 3];
  void pushup(int o) {
    sum[o] = sum[o << 1] + sum[o << 1 | 1];
    if (setv[o << 1] == setv[o << 1 | 1])
      setv[o] = setv[o << 1];
    else
      setv[o] = 0;
  }
  void build(int o, int l, int r) {
    sum[o] = val[o] = setv[o] = 0;
    if (l == r) {
      setv[o] = l;
      return;
    }
    int mid = (l + r) / 2;
    build(o << 1, l, mid);
    build(o << 1 | 1, mid + 1, r);
    pushup(o);
  }
  void pushdown(int o, int l, int r) {
    if (val[o]) {
      int mid = (l + r) >> 1;
      sum[o << 1] += val[o] * (mid - l + 1);
      sum[o << 1 | 1] += val[o] * (r - mid);
      val[o << 1] += val[o];
      val[o << 1 | 1] += val[o];
      setv[o << 1] = setv[o << 1 | 1] = setv[o];
      val[o] = setv[o] = 0;
    }
  }
  void update(int o, int l, int r, int q1, int q2, long long x) {
    if (q1 <= l && r <= q2) {
      if (setv[o]) {
        sum[o] += (r - l + 1) * abs(setv[o] - x);
        val[o] += abs(setv[o] - x);
        setv[o] = x;
        return;
      }
    }
    int mid = (l + r) >> 1;
    pushdown(o, l, r);
    if (q1 <= mid) update(o << 1, l, mid, q1, q2, x);
    if (q2 > mid) update(o << 1 | 1, mid + 1, r, q1, q2, x);
    pushup(o);
  }
  long long query(int o, int l, int r, int q1, int q2) {
    int mid = ((l + r) / 2);
    if (q1 <= l && r <= q2) return sum[o];
    pushdown(o, l, r);
    long long x = 0, y = 0;
    if (q1 <= mid) x = query(o << 1, l, mid, q1, q2);
    if (q2 > mid) y = query(o << 1 | 1, mid + 1, r, q1, q2);
    pushup(o);
    return x + y;
  }
} tree;
int main() {
  int op, a, b, v;
  scanf("%d%d", &n, &m);
  tree.build(1, 1, n);
  while (m--) {
    scanf("%d%d%d", &op, &a, &b);
    if (op == 1) {
      scanf("%d", &v);
      tree.update(1, 1, n, a, b, v);
    } else
      cout << tree.query(1, 1, n, a, b) << endl;
  }
  return 0;
}
