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 ST {
  long long l, r, col, val, pd;
  bool sm;
} segt[1000010];
long long n;
void build_tree(long long x, long long l, long long r) {
  if (l > r) {
    return;
  }
  segt[x].l = l;
  segt[x].r = r;
  segt[x].val = 0;
  segt[x].pd = 0;
  segt[x].sm = false;
  segt[x].col = -1;
  if (l == r) {
    segt[x].sm = true;
    segt[x].col = l;
    return;
  }
  long long mid = (l + r) >> 1;
  build_tree((x << (long long)1), l, mid);
  build_tree((x << (long long)1) + 1, mid + 1, r);
  return;
}
void push_down(long long x) {
  long long l = segt[x].l, r = segt[x].r;
  segt[x].val += segt[x].pd * (r - l + 1);
  segt[(x << (long long)1)].pd += segt[x].pd;
  segt[(x << (long long)1) + 1].pd += segt[x].pd;
  segt[x].pd = 0;
  if (segt[x].sm) {
    segt[(x << (long long)1)].sm = true;
    segt[(x << (long long)1) + 1].sm = true;
    segt[(x << (long long)1)].col = segt[x].col;
    segt[(x << (long long)1) + 1].col = segt[x].col;
  }
  return;
}
void update(long long x, long long l, long long r, long long v) {
  long long tl, tr, mid = (l + r) >> 1, tv;
  tl = segt[x].l;
  tr = segt[x].r;
  if (tl > tr) {
    return;
  }
  push_down(x);
  if (l > tr || r < tl) {
    return;
  }
  tv = abs(segt[x].col - v);
  if (l <= tl && r >= tr) {
    segt[x].col = v;
    if (segt[x].sm) {
      segt[x].pd += tv;
      push_down(x);
      return;
    } else {
      segt[x].sm = true;
    }
  } else {
    if (segt[x].col != v) {
      segt[x].sm = 0;
      segt[x].col = -1;
    }
  }
  update((x << (long long)1), l, r, v);
  update((x << (long long)1) + 1, l, r, v);
  segt[x].val =
      segt[(x << (long long)1)].val + segt[(x << (long long)1) + 1].val;
  return;
}
long long getans(long long x, long long l, long long r) {
  long long tl, tr, mid = (l + r) >> 1;
  tl = segt[x].l;
  tr = segt[x].r;
  if (tl > tr) {
    return 0;
  }
  push_down(x);
  if (l > tr || r < tl) {
    return 0;
  }
  if (l <= tl && r >= tr) {
    return segt[x].val;
  }
  return getans((x << (long long)1), l, r) +
         getans((x << (long long)1) + 1, l, r);
}
int main() {
  long long i, QQ, tp, l, r, x;
  scanf("%I64d%I64d", &n, &QQ);
  build_tree(1, 0, n - 1);
  while (QQ--) {
    scanf("%I64d%I64d%I64d", &tp, &l, &r);
    l--;
    r--;
    if (tp == 1) {
      scanf("%I64d", &x);
      x--;
      update(1, l, r, x);
    } else {
      printf("%I64d\n", getans(1, l, r));
    }
  }
  return 0;
}
