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 int SZ = 300;
int N;
long long add[100005], gadd[100005], gsum[100005];
int col[100005], gcol[100005];
void down(int g) {
  for (int i = g * SZ; i < (g + 1) * SZ; ++i) col[i] = gcol[g];
  gcol[g] = 0;
}
void updateUnit(int l, int x) {
  int g = l / SZ;
  if (gcol[g] != 0) {
    down(g);
    int t = abs(x - col[l]);
    add[l] += t;
    gsum[g] += t;
    col[l] = x;
  } else {
    int t = abs(x - col[l]);
    add[l] += t;
    gsum[g] += t;
    col[l] = x;
  }
}
void updateGroup(int g, int x) {
  if (gcol[g] == 0) {
    for (int i = SZ * g; i < SZ * (g + 1); ++i) {
      int t = abs(x - col[i]);
      add[i] += t;
      gsum[g] += t;
    }
    gcol[g] = x;
  } else {
    int t = abs(x - gcol[g]);
    gadd[g] += t;
    gsum[g] += (long long)SZ * t;
    gcol[g] = x;
  }
}
void update(int l, int r, int x) {
  while (l <= r && l % SZ != 0) {
    updateUnit(l, x);
    ++l;
  }
  while (l + SZ <= r) {
    updateGroup(l / SZ, x);
    l += SZ;
  }
  while (l <= r) {
    updateUnit(l, x);
    ++l;
  }
}
long long query(int l, int r) {
  long long ret = 0;
  while (l <= r && l % SZ != 0) {
    ret += gadd[l / SZ] + add[l];
    ++l;
  }
  while (l + SZ <= r) {
    ret += gsum[l / SZ];
    l += SZ;
  }
  while (l <= r) {
    ret += gadd[l / SZ] + add[l];
    ++l;
  }
  return ret;
}
int main() {
  int M;
  scanf("%d%d", &N, &M);
  for (int i = 0; i < N; ++i) col[i] = i + 1;
  while (M--) {
    int type, x, y;
    scanf("%d%d%d", &type, &x, &y);
    if (type == 1) {
      int z;
      scanf("%d", &z);
      update(x - 1, y - 1, z);
    } else {
      cout << query(x - 1, y - 1) << endl;
    }
  }
  return 0;
}
