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 N, M;
long long Arb[800002], Value[100002], Lazy[800002], Colour[800002];
void Build(long long K, long long L, long long R) {
  long long Mid = (L + R) / 2;
  if (L == R) {
    Colour[K] = L;
    return;
  }
  Build(2 * K, L, Mid);
  Build(2 * K + 1, Mid + 1, R);
}
void Read() { cin >> N >> M; }
void Relax(long long K, long long L, long long R) {
  if (Lazy[K] == 0) return;
  Colour[2 * K] = Colour[2 * K + 1] = Colour[K];
  Arb[2 * K] += Lazy[K] * ((L + R) / 2 - L + 1);
  Arb[2 * K + 1] += Lazy[K] * (R - (L + R) / 2);
  if (L != R) {
    Lazy[2 * K] += Lazy[K];
    Lazy[2 * K + 1] += Lazy[K];
  }
  Lazy[K] = 0;
}
void Update(long long K, long long L, long long R, long long Poz1,
            long long Poz2, long long Val) {
  long long Mid = (L + R) / 2;
  Relax(K, L, R);
  if (L > Poz2 || R < Poz1 || K > 4 * N) return;
  if (Poz1 <= L && R <= Poz2) {
    if (Colour[K] > 0) {
      Arb[K] += abs(Colour[K] - Val) * (R - L + 1);
      Lazy[K] += abs(Colour[K] - Val);
      Colour[K] = Val;
    } else {
      Update(K * 2, L, Mid, Poz1, Poz2, Val);
      Update(K * 2 + 1, Mid + 1, R, Poz1, Poz2, Val);
      Arb[K] = Arb[2 * K] + Arb[2 * K + 1];
      Colour[K] = Val;
    }
    return;
  }
  Update(2 * K + 1, Mid + 1, R, Poz1, Poz2, Val);
  Update(2 * K, L, Mid, Poz1, Poz2, Val);
  Arb[K] = Arb[2 * K] + Arb[2 * K + 1];
  Colour[K] = 0;
}
long long Qwery(long long K, long long L, long long R, long long Poz1,
                long long Poz2) {
  Relax(K, L, R);
  long long Mid = (L + R) / 2;
  if (L > Poz2 || R < Poz1) return 0;
  if (Poz1 <= L && R <= Poz2) return Arb[K];
  return Qwery(2 * K, L, Mid, Poz1, Poz2) +
         Qwery(2 * K + 1, Mid + 1, R, Poz1, Poz2);
}
void Browse() {
  long long i;
  for (i = 1; i <= M; i++) {
    long long type, a, b, c;
    cin >> type >> a >> b;
    if (type == 1) {
      cin >> c;
      Update(1, 1, N, a, b, c);
    }
    if (type == 2) cout << Qwery(1, 1, N, a, b) << "\n";
  }
}
int main() {
  Read();
  Build(1, 1, N);
  Browse();
  return 0;
}
