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-9;
const long long mod = 1e9 + 7;
const int max_n = 101000;
int n, m;
long long input[max_n];
long long seg[4 * max_n];
long long ness[4 * max_n];
long long lazy[4 * max_n];
void construct(int lo = 1, int hi = n, int id = 1) {
  if (lo == hi) {
    seg[id] = lo;
    return;
  }
  int mid = (lo + hi) / 2;
  construct(lo, mid, 2 * id);
  construct(mid + 1, hi, 2 * id + 1);
}
void push_down(int lo, int hi, int id) {
  if (lazy[id] == 0) return;
  int mid = (lo + hi) / 2;
  ness[id * 2] += 1ll * (mid - lo + 1) * lazy[id];
  ness[id * 2 + 1] += 1ll * (hi - mid) * lazy[id];
  seg[2 * id] = seg[2 * id + 1] = seg[id];
  lazy[2 * id] += lazy[id];
  lazy[2 * id + 1] += lazy[id];
  lazy[id] = 0;
}
void modify(int l, int r, int lo, int hi, int id, long long x) {
  if (r < lo || l > hi) return;
  if (lo >= l && hi <= r) {
    if (seg[id] != 0) {
      ness[id] += 1ll * (hi - lo + 1) * abs(x - seg[id]);
      lazy[id] += abs(x - seg[id]);
      seg[id] = x;
      return;
    }
    int mid = (lo + hi) / 2;
    modify(l, r, lo, mid, 2 * id, x);
    modify(l, r, mid + 1, hi, 2 * id + 1, x);
    seg[id] = x;
    ness[id] = ness[id * 2] + ness[id * 2 + 1];
    return;
  }
  push_down(lo, hi, id);
  int mid = (lo + hi) / 2;
  modify(l, r, lo, mid, 2 * id, x);
  modify(l, r, mid + 1, hi, 2 * id + 1, x);
  seg[id] = (seg[id * 2] == seg[id * 2 + 1]) ? seg[id * 2] : 0;
  ness[id] = ness[id * 2] + ness[id * 2 + 1];
}
long long query(int l, int r, int lo = 1, int hi = n, int id = 1) {
  if (r < lo || l > hi) return 1ll * 0;
  if (lo >= l && hi <= r) return ness[id];
  int mid = (lo + hi) / 2;
  push_down(lo, hi, id);
  return query(l, r, lo, mid, 2 * id) + query(l, r, mid + 1, hi, 2 * id + 1);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) input[i] = i;
  for (int i = 1; i <= 4 * n; i++) seg[i] = ness[i] = lazy[i] = 0;
  construct();
  int t, l, r;
  long long x;
  for (int i = 0; i < m; i++) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d %d %I64d", &l, &r, &x);
      modify(l, r, 1, n, 1, x);
    } else {
      scanf("%d %d", &l, &r);
      printf("%I64d\n", query(l, r));
    }
  }
}
