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 MAXN = 101 * 1000;
int col[MAXN * 4], modc[MAXN * 4];
long long sum[MAXN * 4], mods[MAXN * 4];
int len;
void push(int v, int cur_len) {
  if (v < len) {
    if (modc[v] != -1) {
      modc[2 * v + 1] = modc[2 * v] = modc[v];
    }
    mods[2 * v] += mods[v];
    mods[2 * v + 1] += mods[v];
  }
  sum[v] += cur_len * mods[v];
  mods[v] = 0;
  if (modc[v] != -1) {
    col[v] = modc[v];
  }
  modc[v] = -1;
}
void up(int v, int cur_len) {
  if (v < len) {
    push(2 * v, cur_len / 2);
    push(2 * v + 1, cur_len / 2);
    sum[v] = sum[2 * v] + sum[2 * v + 1];
    if (col[2 * v] == col[2 * v + 1] && col[2 * v] != -1) {
      col[v] = col[2 * v];
    } else {
      col[v] = -1;
    }
  }
}
void update(int tl, int tr, int v, int l, int r, int x) {
  if (tl > tr) return;
  push(v, r - l + 1);
  if (tl == l && tr == r && col[v] != -1) {
    mods[v] += abs(x - col[v]);
    modc[v] = x;
    return;
  }
  int s = (l + r) / 2;
  update(tl, min(s, tr), 2 * v, l, s, x);
  update(max(tl, s + 1), tr, 2 * v + 1, s + 1, r, x);
  up(v, r - l + 1);
}
long long get_sum(int tl, int tr, int v, int l, int r) {
  if (tl > tr) return 0;
  push(v, r - l + 1);
  if (tl == l && tr == r) {
    return sum[v];
  }
  int s = (l + r) / 2;
  long long ans = get_sum(tl, min(s, tr), 2 * v, l, s) +
                  get_sum(max(s + 1, tl), tr, 2 * v + 1, s + 1, r);
  up(v, r - l + 1);
  return ans;
}
void dfs(int v, int cur_len) {
  if (v >= len) return;
  dfs(2 * v, cur_len / 2);
  dfs(2 * v + 1, cur_len / 2);
  up(v, cur_len);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (len = 1; len < n; len *= 2)
    ;
  memset(col, -1, sizeof col);
  memset(modc, -1, sizeof modc);
  for (int i = 0; i < n; i++) {
    col[i + len] = i;
    mods[i + len] = 0;
    modc[i + len] = -1;
    sum[i + len] = 0;
  }
  dfs(1, len);
  for (int i = 0; i < m; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r, x;
      cin >> l >> r >> x;
      l--, r--, x--;
      update(l + len, r + len, 1, len, len * 2 - 1, x);
    }
    if (t == 2) {
      int l, r;
      cin >> l >> r;
      l--, r--;
      cout << get_sum(l + len, r + len, 1, len, len * 2 - 1) << '\n';
    }
  }
  return 0;
}
