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;
int n, m;
int a, b, c, d;
struct Tree {
  int l, r, c;
  long long sum, add;
} tree[100005 << 2];
void build(int l, int r, int rt) {
  tree[rt].l = l;
  tree[rt].r = r;
  tree[rt].c = 0;
  tree[rt].sum = 0;
  tree[rt].add = 0;
  if (l == r) {
    tree[rt].c = l;
    return;
  }
  int mid = ((l + r) >> 1);
  build(l, mid, (rt << 1));
  build(mid + 1, r, (rt << 1 | 1));
}
void push_down(int rt) {
  if (tree[rt].c != 0) {
    tree[(rt << 1)].c = tree[(rt << 1 | 1)].c = tree[rt].c;
    tree[rt].c = 0;
  }
  if (tree[rt].add != 0) {
    tree[(rt << 1)].add += tree[rt].add;
    tree[(rt << 1 | 1)].add += tree[rt].add;
    tree[(rt << 1)].sum +=
        (long long)tree[rt].add *
        (long long)(tree[(rt << 1)].r - tree[(rt << 1)].l + 1);
    tree[(rt << 1 | 1)].sum +=
        (long long)tree[rt].add *
        (long long)(tree[(rt << 1 | 1)].r - tree[(rt << 1 | 1)].l + 1);
    tree[rt].add = 0;
  }
}
void push_up(int rt) {
  tree[rt].sum = tree[(rt << 1)].sum + tree[(rt << 1 | 1)].sum;
  if (tree[(rt << 1)].c == tree[(rt << 1 | 1)].c) {
    tree[rt].c = tree[(rt << 1)].c;
    tree[(rt << 1)].c = tree[(rt << 1 | 1)].c = 0;
  } else
    tree[rt].c = 0;
}
void update(int x, int y, int v, int rt) {
  int l = tree[rt].l;
  int r = tree[rt].r;
  if (l >= x && r <= y) {
    if (tree[rt].c != 0) {
      tree[rt].add += abs(v - tree[rt].c);
      tree[rt].sum += (long long)(r - l + 1) * (long long)abs(v - tree[rt].c);
      tree[rt].c = v;
      return;
    }
  }
  push_down(rt);
  int mid = ((l + r) >> 1);
  if (y <= mid)
    update(x, y, v, (rt << 1));
  else if (x >= mid + 1)
    update(x, y, v, (rt << 1 | 1));
  else {
    update(x, mid, v, (rt << 1));
    update(mid + 1, y, v, (rt << 1 | 1));
  }
  push_up(rt);
}
long long query(int x, int y, int rt) {
  int l = tree[rt].l;
  int r = tree[rt].r;
  if (x <= l && y >= r) {
    return tree[rt].sum;
  }
  push_down(rt);
  long long ret = 0;
  int mid = ((l + r) >> 1);
  if (y <= mid)
    ret = query(x, y, (rt << 1));
  else if (x >= mid + 1)
    ret = query(x, y, (rt << 1 | 1));
  else {
    ret += query(x, mid, (rt << 1));
    ret += query(mid + 1, y, (rt << 1 | 1));
  }
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  build(1, n, 1);
  while (m--) {
    scanf("%d", &a);
    if (a == 1) {
      scanf("%d%d%d", &b, &c, &d);
      update(b, c, d, 1);
    } else {
      scanf("%d%d", &b, &c);
      printf("%lld\n", query(b, c, 1));
    }
  }
  return 0;
}
