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;
int n, m;
long long T[400010], d[400010];
map<int, long long> m1;
inline void push(int id, int l, int r) {
  int mid = l + r >> 1;
  d[id << 1] += d[id];
  d[id << 1 | 1] += d[id];
  T[id << 1] += d[id] * (mid - l);
  T[id << 1 | 1] += d[id] * (r - mid);
  d[id] = 0;
  return;
}
long long ask(int x, int y, int id = 1, int l = 1, int r = n + 1) {
  if (l >= y || x >= r) return 0;
  if (x <= l && r <= y) return T[id];
  push(id, l, r);
  return ask(x, y, id << 1, l, l + r >> 1) +
         ask(x, y, id << 1 | 1, l + r >> 1, r);
}
void upd(int x, int y, long long val, int id = 1, int l = 1, int r = n + 1) {
  if (l >= y || x >= r) return;
  if (x <= l && r <= y) {
    T[id] += val * (r - l);
    d[id] += val;
    return;
  }
  push(id, l, r);
  upd(x, y, val, id << 1, l, l + r >> 1);
  upd(x, y, val, id << 1 | 1, l + r >> 1, r);
  T[id] = T[id << 1] + T[id << 1 | 1];
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) m1[i] = i;
  m1[n + 1] = 0;
  for (int i = 0, l, r, x, op; i < m; i++) {
    cin >> op >> l >> r;
    if (op == 1) {
      cin >> x;
      auto it = --m1.upper_bound(r + 1);
      m1[r + 1] = it->second;
      auto rit = m1.lower_bound(r + 1);
      it = --m1.upper_bound(l);
      m1[l] = it->second;
      auto lit = it = m1.lower_bound(l);
      for (auto nit = it; it->first <= r; it++) {
        nit++;
        upd(it->first, nit->first, abs(x - it->second));
      }
      m1.erase(++lit, rit);
      m1[l] = x;
    } else if (op == 2)
      cout << ask(l, r + 1) << endl;
    else
      assert(false);
  }
  return 0;
}
