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 ta[100100], tb[100100], a, b;
void add(long long idx, long long a, long long b) {
  while (idx < 100100) {
    ta[idx] += a;
    tb[idx] += b;
    idx += (idx & -idx);
  }
}
long long get(long long x) {
  a = b = 0;
  long long idx = x;
  while (idx > 0) {
    a += ta[idx];
    b += tb[idx];
    idx -= (idx & -idx);
  }
  return a * x + b;
}
void update(long long l, long long r, long long v) {
  add(l, v, -l * v + v);
  add(r + 1, -v, (r - l + 1) * v + l * v - v);
}
long long query(long long l, long long r) { return get(r) - get(l - 1); }
long long N, Q, op, l, r, col, cnt;
struct iv {
  long long l, r;
  long long col;
  iv() {}
  iv(long long _l, long long _r, long long _col) {
    l = _l;
    r = _r;
    col = _col;
  }
  bool operator<(iv o) const { return l < o.l; }
} curr[100100];
set<iv> S;
set<iv>::iterator lo, hi, it;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> Q;
  for (long long I = 1; I <= N; I++) S.insert(iv(I, I, I));
  while (Q--) {
    cin >> op >> l >> r;
    if (op == 1) {
      cin >> col;
      lo = S.upper_bound(iv(l, 0, 0));
      hi = S.upper_bound(iv(r, 0, 0));
      lo--;
      cnt = 0;
      for (it = lo; it != hi; it++) curr[cnt++] = *it;
      S.erase(lo, hi);
      if (curr[0].l < l) {
        S.insert(iv(curr[0].l, l - 1, curr[0].col));
        curr[0].l = l;
      }
      if (curr[cnt - 1].r > r) {
        S.insert(iv(r + 1, curr[cnt - 1].r, curr[cnt - 1].col));
        curr[cnt - 1].r = r;
      }
      S.insert(iv(l, r, col));
      for (long long I = 0; I < cnt; I++)
        update(curr[I].l, curr[I].r, abs(curr[I].col - col));
    } else if (op == 2) {
      cout << query(l, r) << "\n";
    } else
      assert(false);
  }
}
