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 long long MAXN = (1 << 20);
long long myabs(long long x) { return x < 0 ? -x : x; }
struct node {
  long long col, sum, lazy;
  node() {
    sum = 0;
    col = -2;
    lazy = 0;
  }
  node(long long _col, long long _sum) {
    col = _col;
    sum = _sum;
    lazy = 0;
  }
};
node temp, broken;
node merge(node l, node r) {
  if (l.col == -2) return r;
  if (r.col == -2) return l;
  temp.lazy = 0;
  temp.col = (l.col == r.col) ? l.col : -1;
  temp.sum = l.sum + r.sum;
  return temp;
}
struct segment_tree {
  node tr[4 * MAXN];
  void push(long long l, long long r, long long idx) {
    tr[idx].sum += tr[idx].lazy * (r - l + 1);
    if (l != r) {
      tr[2 * idx + 1].lazy += tr[idx].lazy;
      tr[2 * idx + 2].lazy += tr[idx].lazy;
      if (tr[idx].col != -1) {
        tr[2 * idx + 1].col = tr[idx].col;
        tr[2 * idx + 2].col = tr[idx].col;
      }
    }
    tr[idx].lazy = 0;
  }
  void init(long long l, long long r, long long idx) {
    if (l == r) {
      tr[idx] = node(l, 0);
      return;
    }
    long long mid = (l + r) >> 1;
    init(l, mid, 2 * idx + 1);
    init(mid + 1, r, 2 * idx + 2);
    tr[idx] = merge(tr[2 * idx + 1], tr[2 * idx + 2]);
  }
  void update(long long qL, long long qR, long long nw_col, long long l,
              long long r, long long idx) {
    push(l, r, idx);
    if (qL <= l && r <= qR && tr[idx].col != -1) {
      tr[idx].lazy += myabs(nw_col - tr[idx].col);
      tr[idx].col = nw_col;
      push(l, r, idx);
      return;
    }
    if (qL > r || l > qR) return;
    long long mid = (l + r) >> 1;
    update(qL, qR, nw_col, l, mid, 2 * idx + 1);
    update(qL, qR, nw_col, mid + 1, r, 2 * idx + 2);
    tr[idx] = merge(tr[2 * idx + 1], tr[2 * idx + 2]);
  }
  node query(long long qL, long long qR, long long l, long long r,
             long long idx) {
    push(l, r, idx);
    if (qL <= l && r <= qR) return tr[idx];
    if (qL > r || l > qR) return broken;
    long long mid = (l + r) >> 1;
    return merge(query(qL, qR, l, mid, 2 * idx + 1),
                 query(qL, qR, mid + 1, r, 2 * idx + 2));
  }
};
long long n, m;
void read() { cin >> n >> m; }
segment_tree t;
void solve() {
  t.init(1, n, 0);
  while (m--) {
    long long type;
    cin >> type;
    if (type == 1) {
      long long l, r, val;
      cin >> l >> r >> val;
      t.update(l, r, val, 1, n, 0);
    } else {
      long long l, r;
      cin >> l >> r;
      cout << t.query(l, r, 1, n, 0).sum << '\n';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}
