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;
template <typename T, typename L>
bool smax(T &x, L y) {
  return x < y ? (x = y, true) : false;
}
template <typename T, typename L>
bool smin(T &x, L y) {
  return y < x ? (x = y, true) : false;
}
void upd(long long, long long);
void upd1(long long, long long);
const long long MAXN = 1e5 + 15;
const long long SQ = 325;
long long n, m, tme;
pair<long long, long long> info[MAXN], grinfo[SQ];
bool rel[SQ];
long long ans1[SQ], ans2[SQ], ans3[MAXN];
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cerr << "HELLO WORLD :)\n";
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) info[i] = {tme, i};
  while (m--) {
    tme++;
    long long t, l, r;
    cin >> t >> l >> r;
    r++;
    long long gl = l / SQ;
    long long gr = r / SQ;
    if (t == 1) {
      long long c;
      cin >> c;
      for (long long i = l; i < r;) {
        long long gp = i / SQ;
        if (gp > gl && gp < gr) {
          upd1(gp, c);
          i += SQ;
        } else
          upd(i++, c);
      }
    } else {
      long long ret = 0;
      for (long long i = l; i < r;) {
        long long gp = i / SQ;
        if (gp > gl && gp < gr) {
          ret += ans1[gp];
          i += SQ;
        } else
          ret += ans2[gp] + ans3[i++];
      }
      cout << ret << '\n';
    }
  }
  return false;
}
void upd(long long p, long long c) {
  long long gp = p / SQ, col = max(info[p], grinfo[p / SQ]).second;
  ans3[p] += abs(col - c);
  ans1[gp] += abs(col - c);
  rel[gp] = 0;
  info[p] = {tme, c};
}
void upd1(long long gp, long long col) {
  long long c = grinfo[gp].second;
  if (rel[gp]) {
    ans1[gp] += abs(col - c) * SQ;
    ans2[gp] += abs(col - c);
  } else
    for (long long i = gp * SQ; i < (gp + 1) * SQ; i++) upd(i, col);
  grinfo[gp] = {tme, col};
  rel[gp] = true;
}
