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;
long long power(long long a, long long b, long long m) {
  long long x = 1 % m;
  a %= m;
  while (b) {
    if ((1 & b)) x = x * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return x;
}
struct node {
  int all_equal;
  long long delta;
  long long answer;
  node() { all_equal = delta = answer = 0; }
} seg[100005 << 2];
void build(int t, int i, int j) {
  if (i == j) {
    seg[t].all_equal = i;
    return;
  }
  int left = t << 1, right = left | 1, mid = (i + j) >> 1;
  build(left, i, mid);
  build(right, mid + 1, j);
}
int val;
void push(int t, int i, int j) {
  if (seg[t].all_equal) {
    seg[t].delta += abs(seg[t].all_equal - val);
    seg[t].answer += abs(seg[t].all_equal - val) * 1LL * (j - i + 1);
    seg[t].all_equal = 0;
    return;
  }
  int left = t << 1, right = left | 1, mid = (i + j) >> 1;
  push(left, i, mid);
  push(right, mid + 1, j);
  seg[t].answer =
      seg[left].answer + seg[right].answer + seg[t].delta * (j - i + 1);
}
void update(int t, int i, int j, const int &ri, const int &rj) {
  if (ri <= i && j <= rj) {
    push(t, i, j);
    seg[t].all_equal = val;
    return;
  }
  int left = t << 1, right = left | 1, mid = (i + j) >> 1;
  if (seg[t].all_equal) {
    seg[left].all_equal = seg[right].all_equal = seg[t].all_equal;
    seg[t].all_equal = 0;
  }
  if (ri <= mid) update(left, i, mid, ri, rj);
  if (rj > mid) update(right, mid + 1, j, ri, rj);
  seg[t].answer =
      seg[left].answer + seg[right].answer + seg[t].delta * (j - i + 1);
}
long long query(int t, int i, int j, int ri, const int rj) {
  if (ri <= i && j <= rj) {
    return seg[t].answer;
  }
  long long l = 0, r = 0;
  int left = t << 1, right = left | 1, mid = (i + j) >> 1;
  if (ri <= mid) {
    int ind = min(rj, mid);
    l = query(left, i, mid, ri, ind) + seg[t].delta * (ind - ri + 1);
  }
  if (rj > mid) {
    int ind = max(ri, mid + 1);
    r = query(right, mid + 1, j, ind, rj) + seg[t].delta * (rj - ind + 1);
  }
  return l + r;
}
int main() {
  int n, m;
  scanf("%d", &n);
  scanf("%d", &m);
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int t, l, r;
    scanf("%d", &t);
    scanf("%d", &l);
    scanf("%d", &r);
    if (t == 1) {
      scanf("%d", &val);
      update(1, 1, n, l, r);
    } else {
      long long ans = query(1, 1, n, l, r);
      printf("%lld\n", ans);
    }
  }
}
