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;
  int r;
  long long val;
  long long clr;
  long long laz;
};
node tree[400010];
int n, q;
void pushup(int cur) {
  tree[cur].val = tree[2 * cur].val + tree[2 * cur + 1].val;
  if (tree[2 * cur].clr != 0 && tree[2 * cur + 1].clr != 0 &&
      tree[2 * cur].clr == tree[2 * cur + 1].clr) {
    tree[cur].clr = tree[2 * cur].clr;
  } else
    tree[cur].clr = 0;
}
void pushdown(int cur) {
  if (tree[cur].laz != 0) {
    tree[2 * cur].val +=
        (tree[2 * cur].r - tree[2 * cur].l + 1) * tree[cur].laz;
    tree[2 * cur].clr = tree[cur].clr;
    tree[2 * cur].laz += tree[cur].laz;
    tree[2 * cur + 1].val +=
        (tree[2 * cur + 1].r - tree[2 * cur + 1].l + 1) * tree[cur].laz;
    tree[2 * cur + 1].clr = tree[cur].clr;
    tree[2 * cur + 1].laz += tree[cur].laz;
    tree[cur].laz = 0;
  }
}
void build(int l, int r, int cur) {
  int m;
  tree[cur].l = l;
  tree[cur].r = r;
  tree[cur].val = 0;
  tree[cur].clr = 0;
  tree[cur].laz = 0;
  if (l == r) {
    tree[cur].clr = l;
    return;
  }
  m = (l + r) / 2;
  build(l, m, 2 * cur);
  build(m + 1, r, 2 * cur + 1);
  pushup(cur);
}
void update(int pl, int pr, long long clr, int cur) {
  if (pl <= tree[cur].l && tree[cur].r <= pr && tree[cur].clr != 0) {
    tree[cur].val +=
        (tree[cur].r - tree[cur].l + 1) * (long long)abs(tree[cur].clr - clr);
    tree[cur].laz += (long long)abs(tree[cur].clr - clr);
    tree[cur].clr = clr;
    return;
  }
  pushdown(cur);
  if (pl <= tree[2 * cur].r) update(pl, pr, clr, 2 * cur);
  if (pr >= tree[2 * cur + 1].l) update(pl, pr, clr, 2 * cur + 1);
  pushup(cur);
}
long long query(int pl, int pr, int cur) {
  long long res;
  if (pl <= tree[cur].l && tree[cur].r <= pr) {
    return tree[cur].val;
  }
  pushdown(cur);
  res = 0;
  if (pl <= tree[2 * cur].r) res += query(pl, pr, 2 * cur);
  if (pr >= tree[2 * cur + 1].l) res += query(pl, pr, 2 * cur + 1);
  return res;
}
int main() {
  long long clr;
  int i, op, l, r;
  scanf("%d%d", &n, &q);
  build(1, n, 1);
  while (q--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d%lld", &l, &r, &clr);
      update(l, r, clr, 1);
    } else {
      scanf("%d%d", &l, &r);
      printf("%lld\n", query(l, r, 1));
    }
  }
  return 0;
}
