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;
inline int Mid(int a, int b) { return (a + b) >> 1; }
inline long long ABS(long long x) { return x > 0 ? x : -x; }
int n, m;
struct node {
  int l, r, id;
  long long siz() { return (long long)(r - l + 1); }
  long long sum, col, lazy, same;
} tree[100005 << 2];
void push_down(int id) {
  if (tree[id].l == tree[id].r) return;
  if (tree[id].lazy) {
    tree[(id << 1)].sum += tree[(id << 1)].siz() * tree[id].lazy;
    tree[(id << 1 | 1)].sum += tree[(id << 1 | 1)].siz() * tree[id].lazy;
    tree[(id << 1)].lazy += tree[id].lazy;
    tree[(id << 1 | 1)].lazy += tree[id].lazy;
    tree[id].lazy = 0;
  }
  if (tree[id].col >= 0) {
    tree[(id << 1)].col = tree[(id << 1 | 1)].col = tree[id].col;
  }
}
void push_up(int id) {
  if (tree[(id << 1)].col == tree[(id << 1 | 1)].col &&
      tree[(id << 1)].col >= 0)
    tree[id].col = tree[(id << 1)].col, tree[id].same = 1;
  else
    tree[id].col = -1, tree[id].same = 0;
  tree[id].sum = tree[(id << 1)].sum + tree[(id << 1 | 1)].sum;
}
void build(int l, int r, int id) {
  tree[id].l = l, tree[id].r = r;
  tree[id].sum = 0;
  tree[id].lazy = 0;
  tree[id].col = -1;
  tree[id].same = 0;
  if (l == r) {
    tree[id].col = l;
    tree[id].same = 1;
    return;
  }
  int mid = Mid(l, r);
  build(l, mid, (id << 1));
  build(mid + 1, r, (id << 1 | 1));
}
void updata(int l, int r, int id, long long col) {
  push_down(id);
  if (l == tree[id].l && tree[id].r == r && tree[id].same) {
    tree[id].sum += tree[id].siz() * ABS(col - tree[id].col);
    tree[id].lazy += ABS(col - tree[id].col);
    tree[id].col = col;
    return;
  }
  int mid = Mid(tree[id].l, tree[id].r);
  if (mid < l)
    updata(l, r, (id << 1 | 1), col);
  else if (r <= mid)
    updata(l, r, (id << 1), col);
  else {
    updata(l, mid, (id << 1), col);
    updata(mid + 1, r, (id << 1 | 1), col);
  }
  push_up(id);
}
long long query(int l, int r, int id) {
  push_down(id);
  if (l == tree[id].l && tree[id].r == r) return tree[id].sum;
  int mid = Mid(tree[id].l, tree[id].r);
  if (mid < l)
    return query(l, r, (id << 1 | 1));
  else if (r <= mid)
    return query(l, r, (id << 1));
  else
    return query(l, mid, (id << 1)) + query(mid + 1, r, (id << 1 | 1));
}
int main() {
  int type, l, r;
  long long x;
  while (~scanf("%d %d", &n, &m)) {
    build(1, n, 1);
    while (m--) {
      scanf("%d %d %d", &type, &l, &r);
      if (type == 1) {
        scanf("%I64d", &x);
        updata(l, r, 1, x);
      } else {
        printf("%I64d\n", query(l, r, 1));
      }
    }
  }
  return 0;
}
