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, r;
  long long t, c, x;
} t[800010];
int n, m;
void build(int l, int r, int n) {
  if (l == r)
    t[n] = (node){l, l, 0, l, 0};
  else {
    int mid = (l + r) >> 1;
    t[n] = (node){l, r, 0, 0, 0};
    build(l, mid, (n << 1)), build(mid + 1, r, ((n << 1) | 1));
  }
}
void pluss(int n, long long k) {
  t[n].x += k * (t[n].r - t[n].l + 1), t[n].t += k;
}
void push(int n) {
  if (t[n].t)
    pluss((n << 1), t[n].t), pluss(((n << 1) | 1), t[n].t), t[n].t = 0;
  if (t[n].c) t[(n << 1)].c = t[((n << 1) | 1)].c = t[n].c;
}
void updata(int l, int r, long long c, int n) {
  if ((l <= t[n].l) && (t[n].r <= r) && (t[n].c)) {
    pluss(n, fabs(t[n].c - c));
    t[n].c = c;
  } else {
    push(n);
    int mid = (t[n].l + t[n].r) >> 1;
    if (l <= mid) updata(l, r, c, (n << 1));
    if (mid < r) updata(l, r, c, ((n << 1) | 1));
    t[n].x = t[(n << 1)].x + t[((n << 1) | 1)].x;
    t[n].c = (t[(n << 1)].c == t[((n << 1) | 1)].c) ? (t[(n << 1)].c) : 0;
  }
}
long long query(int l, int r, int n) {
  if ((l <= t[n].l) && (t[n].r <= r))
    return t[n].x;
  else {
    push(n);
    int mid = (t[n].l + t[n].r) >> 1;
    long long res = 0;
    if (l <= mid) res += query(l, r, (n << 1));
    if (mid < r) res += query(l, r, ((n << 1) | 1));
    t[n].x = t[(n << 1)].x + t[((n << 1) | 1)].x;
    t[n].c = (t[(n << 1)].c == t[((n << 1) | 1)].c) ? (t[(n << 1)].c) : 0;
    return res;
  }
}
void work() {
  scanf("%d %d", &n, &m);
  build(1, n, 1);
  for (int i = 1, t, l, r, x; i <= m; i++) {
    scanf("%d", &t);
    if (t == 1)
      scanf("%d %d %d", &l, &r, &x), updata(l, r, x, 1);
    else
      scanf("%d %d", &l, &r), printf("%I64d\n", query(l, r, 1));
  }
}
int main() {
  work();
  return 0;
}
