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 double eps(1e-8);
const double pi(3.14159265358979);
const long long N = 100100;
long long getint() {
  char ch;
  while (!isdigit(ch = getchar()))
    ;
  long long x = ch - '0';
  while (isdigit(ch = getchar())) x = x * 10 + ch - '0';
  return x;
}
struct node {
  long long l, r;
  long long lazy, sum;
  long long same_color, c;
} t[N << 2] = {};
void make_tree(long long l, long long r, long long k) {
  if (l != r) {
    long long mid = (l + r) >> 1;
    make_tree(l, mid, k << 1);
    make_tree(mid + 1, r, (k << 1) + 1);
    t[k].same_color = false;
  } else {
    t[k].same_color = true;
    t[k].c = l;
  }
  t[k].l = l;
  t[k].r = r;
  t[k].lazy = t[k].sum = 0;
}
void set_lazy(long long k, long long lazy) {
  t[k].lazy += lazy;
  t[k].sum += (t[k].r - t[k].l + 1) * lazy;
}
void push(long long k) {
  if (t[k].l != t[k].r) {
    set_lazy(k << 1, t[k].lazy);
    set_lazy((k << 1) + 1, t[k].lazy);
    if (t[k].same_color) {
      t[k << 1].same_color = true;
      t[k << 1].c = t[k].c;
      t[(k << 1) + 1].same_color = true;
      t[(k << 1) + 1].c = t[k].c;
    }
  }
  t[k].lazy = 0;
}
long long sum(long long l, long long r, long long k) {
  if (t[k].l >= l && t[k].r <= r) return t[k].sum;
  push(k);
  long long s = 0;
  if (t[k << 1].r >= l) s += sum(l, r, k << 1);
  if (t[(k << 1) + 1].l <= r) s += sum(l, r, (k << 1) + 1);
  return s;
}
void change(long long l, long long r, long long c, long long k) {
  if (t[k].l >= l && t[k].r <= r && t[k].same_color) {
    set_lazy(k, max(c, t[k].c) - min(c, t[k].c));
    t[k].c = c;
    return;
  }
  push(k);
  if (t[k << 1].r >= l) change(l, r, c, k << 1);
  if (t[(k << 1) + 1].l <= r) change(l, r, c, (k << 1) + 1);
  t[k].sum = t[k << 1].sum + t[(k << 1) + 1].sum;
  if (t[k].l >= l && t[k].r <= r)
    t[k].same_color = true, t[k].c = c;
  else
    t[k].same_color = false;
}
long long n, m;
void init() {
  n = getint(), m = getint();
  make_tree(1, n, 1);
}
void print(long long k) {
  if (t[k].l != t[k].r) push(k);
  if (t[k].l != t[k].r) print(k << 1);
  cout << t[k].l << ' ' << t[k].r << ' ' << t[k].same_color << ' ' << t[k].c
       << ' ' << t[k].sum << endl;
  if (t[k].l != t[k].r) print((k << 1) + 1);
}
void work() {
  long long type, l, r, x;
  for (long long i = 1; i <= m; ++i) {
    type = getint();
    if (type == 1) {
      l = getint(), r = getint(), x = getint();
      change(l, r, x, 1);
    }
    if (type == 2) {
      l = getint(), r = getint();
      printf("%I64d\n", sum(l, r, 1));
    }
  }
}
int main() {
  init();
  work();
  return 0;
}
