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;
struct node {
  int lt, rt, color;
  long long sum, len, add;
} tree[maxn << 2];
void pushup(int v) {
  tree[v].sum = tree[v << 1].sum + tree[v << 1 | 1].sum;
  if (tree[v << 1].color == tree[v << 1 | 1].color)
    tree[v].color = tree[v << 1].color;
  else
    tree[v].color = 0;
}
void pushdown(int v) {
  if (tree[v].add) {
    tree[v << 1].add += tree[v].add;
    tree[v << 1 | 1].add += tree[v].add;
    tree[v << 1].sum += tree[v].add * tree[v << 1].len;
    tree[v << 1 | 1].sum += tree[v].add * tree[v << 1 | 1].len;
    tree[v].add = 0;
  }
  if (tree[v].color) {
    tree[v << 1].color = tree[v << 1 | 1].color = tree[v].color;
    tree[v].color = 0;
  }
}
void build(int lt, int rt, int v) {
  tree[v].lt = lt;
  tree[v].rt = rt;
  tree[v].len = rt - lt + 1;
  tree[v].add = 0;
  tree[v].sum = 0;
  if (lt == rt) {
    tree[v].color = lt;
    return;
  }
  int mid = (lt + rt) >> 1;
  build(lt, mid, v << 1);
  build(mid + 1, rt, v << 1 | 1);
  pushup(v);
}
void update(int lt, int rt, int color, int v) {
  if (lt <= tree[v].lt && rt >= tree[v].rt && tree[v].color) {
    tree[v].add += abs(tree[v].color - color);
    tree[v].sum += abs(tree[v].color - color) * tree[v].len;
    tree[v].color = color;
    return;
  }
  pushdown(v);
  if (lt <= tree[v << 1].rt) update(lt, rt, color, v << 1);
  if (rt >= tree[v << 1 | 1].lt) update(lt, rt, color, v << 1 | 1);
  pushup(v);
}
long long query(int lt, int rt, int v) {
  if (lt <= tree[v].lt && rt >= tree[v].rt) return tree[v].sum;
  pushdown(v);
  long long sum = 0;
  if (lt <= tree[v << 1].rt) sum += query(lt, rt, v << 1);
  if (rt >= tree[v << 1 | 1].lt) sum += query(lt, rt, v << 1 | 1);
  pushup(v);
  return sum;
}
int main() {
  int n, m, op, x, y, color;
  scanf("%d%d", &n, &m);
  build(1, n, 1);
  while (m--) {
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      scanf("%d", &color);
      update(x, y, color, 1);
    } else
      printf("%I64d\n", query(x, y, 1));
  }
  return 0;
}
